Motion in straight Line

Introduction Position and Displacement Average velocity and speed Instantaneous velocity and speed Acceleration Motion with constant acceleration Free fall acceleration Relative velocity Solved Examples Part 1 Solved Examples Part 2 Solved Examples Part 3 Solved Examples Part 4 Solved Examples Part

1. Introduction

In our dialy life ,we see lots of things moving around for example car passing through from one place to other ,person riding on a bicycle and many more like this.

In scientific terms an object is said to be in motion ,if it changes its position with the pasage of time and if it does not change it position with the passage of time then it is said to be at rest

Both the motion and rest are relative terms for example mobile kept on the table is resting at its position but it is moving in the sense as earth is rotating on its axis.So for a person seeing mobile from earth it is at rest and for person on moon earth seems to change its position with time and so mobile is moving.

Simplest case of motion is rectilinear motion which is the motion of the object in a straight line In our descriotion of object ,we will treat the object as an point object Object under consideration can be treated as point object if the size of the object is much

smaller than the distance travelled by it in a reasonable time duration for example length of a motor car travelling a distance of 500km can be neglected w.r.t distance travelled by it.

Here in kinematics ,we study ways to describe the motion without going into the cause of the motion

2.Position and Displacement

(a) Position:

To locate the position in motion or at rest,we need a frame of refrence. Simplest way to choose a frame of refrence is to choose three mutually perpendicular axis

labelled as X-,Y- and Z- axis as shown in figure below

Such system of labelling position of an object is known as rectangular coordinates system If A(x,y,z) be the position of any point in rectangular co-ordinates system it can be labelled as

follows

Point O is the point of intersection of these mutaully perpendicular axis and is known as refrence point or origin of frame of refrence

To measure a time ,we can also attach a clock with this frame of refrence If any or all co-ordinates of the object under consideration changes with time in this frame of

refrence then the object is said to be in a motion w.r.t the frame of the refrence otherwise it is at rest

For describing motion in one dimension we need one set of co-ordinates axis i.e only one of X,Y and Z axis

Similary for two and three dimensions we need two or three set of axis respectively Motion of an object along a straight line is an example of motion in one dimension For such a motion,any one axis say X-axis may be choosen so as to co-incide with the path

along which object is moving Position of the object can be measured w.r.t origin O shown in the figure

Position to the right of the origin has positive values and those to the left of origin O has negative values.

(b) Distance and displacement:

In the graph shown below an object is at position P at time t1 and at position R at time t2.

In the time interval from t1 to t2 particle has travelled path PQR and length of the path PQR is the distance travelled by the object in the time interval t1 to t2

Now connect the initial position of the object P with its final position R through a straight line and we get the displacement of the object.

Displacement of the object has both magnitude and direction i.e., displacement is a vector quantity.

Magnitude of displacement vector is equal to the length of straight line joining initial and final position and its direction points from the initial position of object towards its final position.

In contrast to displacement distance is scalar quantity.

Question:A particle is thrown vertically upwards. It reaches the height H and the come downward.What is the

distance and displacement in the whole motion?For solution click hereQuestion:True and False Statementa) Distance is a Vector quantity?b) Displacement is a Vector quantity?c) Dimension of Distance and displacement are same?d) Displacement on round trip is zero?

3. Average velocity and speed

Consider a particle undergoing motion along a straight line i.e. moving along X-axis. X co-ordinate describing motion of the particle from origin O varies with time or we can say

that X co-ordinate depends on time. If at time t=t1 particleis at point P , at a distance x1 from origin and at time t=t2 it is at point Q at

a distance x2 from the origin then displacement during this time is a vector from point P to Q and isΔx=x2-x1                             (1)

The average velocity of the prticle is defined as the ratio of the displacement Δx of the particle in the time interval Δt=t2-t1. If vavg represents average velocity then,

Figure 5b represents the co-ordinate time graph of the motion of the particle i.e., it shows how the value of x-coordinate of moving particle changes with the passage of time.

In figure 5b average velocity of the particle is represented by the slope of chord PQ which is equal to the ratio of the displacement Δx occuring in the particular time interval Δt.

Like displacement average velocity vavg also has magnitude as well as direction i.e., average velocity is a vector quantity.

Average velocity of the particle can be positive as well as negative and its positive and negative value depends on the sign of displacement.

If displacement of particle is zero its average velocity is also zero. Graphs below shows the x-t graphs of particle moving with positive, negative average velocity

and the particle at rest.

From graph 5c it is clear that for positive average velocity slope of line slants upwards right or we can say that it has positive slope.

For negative average velocity slope line slants upwards down to the right i.e. it has negative slope.

For particles at rest slope is zero. So far we have learnt that average speed is the rate of motion over displacement of the

object. Displacement of the object is different from the actual distance travelled by the particle. For actual distance travelled by the particle its average speed is defined as the total distance

travelled by the particle in the time interval during which the motion takes place. Mathematically,

Since distance travelled by an particle does not involve direction so speed of the particle depending on distance travelled does not involve direction and hence is a scalar quantity and is always positive.

Magnitude of average speed may differ from average velocity because motion in case of average speed involve distance which may be greater than magnitude of displacement for ex.

here a man starts travelling from origin till point Q and return to point P then in this case displacement of man isDisplacement from O to Q is OQ=80mDisplacement from Q to P is =20m-80m=-60mtotal displacement of particle in moving from O to Q and then moving Q to P is = 80m + (-60m) = 20 mNow total distance travelled by man is OQ+OP= 80m +60m = 140mHence diring same cource of motion distance travelled is greater then displacement.

from this we can say that average speed depending on distance is in general greater than magnitude of velocity.

Question:A car moves a distance 100 km in 2 hours. What is the average speed of the car?Question:A man moves from his residence to market and return back to residence in 2 hours. The distance between market and residence is 20 km. a) what is average velocity of the man? b) what is the average speed of the man?

Question:True and False Statementa) Average speed is a scalar quantity?b) Velocity is a Vector quantity?c) Dimension of Velocity is LT-1?

4. Instantaneous velocity and speed

Velocity of particle at any instant of time or at any point of its path is called instantaneous velocity.

Again consider the graph 5b and imagine second point Q being taken more and more closer to point P then calculate the average velocity over such short displacement and time interval.

Instantaneous velocity can be defined as limiting value of average velocity when second point comes closer and closer to the first point.

Limiting value of Δx/ Δt as Δt aproaches zero is written as dx/dt, and is known as instantaneous velocity.Thus instantaneous velocity is

As point Q aproaches point P in figure 5a in this limit slope of the chord PQ becomes equal to the slope of tangent to the curve at point P.

Thus we can say that instantaneous velocity at any point of a coordinate time graph is equal to the slope of the tangent to the graph at that point.

Instantaneous speed or speed is the magnitude of the instantaneous velocity unlike the case of average velocity and average speed where average speed over an finite interval of time may be greate than or equal to average velocity.

Unit of average velocity , average speed, instantaneous velocity and instantaneous speed is ms-1 in SI system of units.

Some other units of velocity are ft.s-1 , cm.s-1 .

5. Acceleration

Acceleration is the rate of change of velocity with time. For describing average acceleration we first consider the motion of an object along X-axis. Suppose at time t1 object is at point P moving with velocity v1 and at time t2 it is at point Q and

has velocity v2. Now average acceleration of object in moving from P to Q is

which is the change in velocity of object with the passage of time.

Instantaneous acceleration can be defined in the same way as instantaneous velocity

The instantaneous acceleration at any instant is the slope of v-t graph at that instant.

In figure 7 instantaneous acceleration at point P is equal to the slope of tangent at this point P.

Since velocity of a moving object has both magnitude and direction likewise acceleration depending on velocity has both magnitude and direction and hence acceleration is a vector quantity.

Acceleration can also be positive, negative and zero. SI unit of acceleration is ms-2

Question:A car moves in a straight road. The velocity at point A is 50 km/hr, It presses the accelerator paddle and reach point B in 10 sec. The velocity ar point B is 100Km/hrFind the average acceleration of the car from point A to point B?

Question:True and False Statementa) Acceleration is a scalar quantity?b) if the velocity of the object is zero,then acceleration will always be zero?c) Instantanous speed is always equal to magnitude of instantaous velocity

6. Motion with constant acceleration

Motion with constant acceleration or uniformly accelerated motion is that in which velocity changes at the same rate troughout motion.

When the acceleration of the moving object is constant its average acceleration and instantaneous acceleration are equal. Thus from eq. 5 we have

Let v0 be the velocity at initial time t=0 and v be the velocity of object at some other instant of time say at t2=t then above eq. 7 becomes

or , v=v0+at (8) Graphically this relation is represented in figure 8 given below.

Thus from the graph it can be seen clearly that velocity v at time t is equal to the velocity v0 at time t=0 plus the change in velocity (at).

In the same way average velocity can be written as

where x0 is the position of object at time t=0 and vavg is the averag velocity between time t=0 to time t.The above equation then givesx=x0+vavgt (9)but for the interval t=0 to t the average velocity is

Now from eq. 8 we findvavg = v0 + ½(at) (11)putting this in eq. 9 we find x = x0 + v0t + ½(at2) or,x - x0 = v0t + ½(at2) (12)this is the position time relation for object having uniformly accelerated motion.

From eq. 12 it is clear that an object at any time t has quadratic dependance on time, when it moves with constant acceleration along a straight line and x-t graph for such motion will be parabolic in natureas shown below.

Equation 8 and 12 are basic equations for constant acceleration and these two equations can be combined to get yet another relation for x , v and a eliminating t so, from 8

putting this value of t in equation 12 and solving it we finally get,v2 = (v0)2 + 2a ( x - x0 ) (13)

Thus from equation 13 we see that it is velocity displacement relation between velocities of object moving with constant acceleration at time t and t=0 and their corresponding positions at these intervals of time.

This relation 13 is helpful when we do not know time t. Likewise we can also eliminate the acceleration between equation 8 and 12. Thus from

equation 8

putting this value of a in equation 12 and solving it we finally get,( x - x0 ) = ½ ( v0 + v ) t (14)

Same way we can also eliminate v0 using equation 8 and 12. Now from equation 8v0 = v - atputting this value of v0 in equation 12 and solving it we finally get,( x - x0 ) = vt + ½ ( at2 ) (15)thus equation 15 does not involve initial velocity v0

Thus these basic equations 8 and 12 , and derived equations 13, 14, and 15 can be used to solve constant acceleration problems.

7. Free fall acceleration

Freely falling motion of any body under the effect of gravity is an example of uniformly accelerated motion.

Kinematic equation of motion under gravity can be obtained by replacing acceleration 'a' in equations of motion by acceleration due to gravity 'g'.

Value of g is equal to 9.8 m.s-2. Thus kinematic equations of motion under gravity are

v = v0 + gt                          (16a)

x = v0t + ½ ( gt2 )          (16b)v2 = (v0)2 + 2gx           (16c)

The value of g is taken positive when the body falls vertically downeards and negative when the body is projected up against gravity.

8. Relative velocity

Consider two objects A and B moving with uniform velocities vAand vB along two straight and parallel tracks.

Let xOAand xOB be their distances from origin at time t=0 and xAand xB be their distances from origin at time t.

For object AxA = xOA + vAt               (18)and for object BxB = xOB + vBt                (19)subtracting equation 18 from 19xB - xA = ( xOB - xOA ) + ( vB - vA) t                          (20)

Above equation 20 tells that as seen from object A , object B seems to have velocity ( vB - vA) .

Thus ( vB - vA ) is the velocity of object B relative to object A. Thus,vBA = ( vB - vA )                (21)

Similarly velocity of object A relative to object B isvAB = ( vA - vB )                (22)

If vB = vA then from equation 20 xB - xA = ( xOB - xOA )i.e., two objects A and B stays apart at constant distance.

vA > vB then (vB - vA ) would be negative and the distance between two objects will go on decreasing by an amount ( vA - vB ) after each unit of time. After some time they will meet and then object A will overtake object B.

If vA and vB have opposite signs then magnitude of vBA or vAB would be greater then the magnitude of velocity of A or that of B and objects seems to move very fast.

Question 1. A ball is dropped vertically from a height h above the ground .It hits the ground and bounces up vertically to a height h/2.Neglecting subsequent motion and air resistance ,its velocity v varies with the height h as

Solution(1):. Before hitting the ground ,the velocity v is given by v2=2gd Which is a quadratic equation and hence parabolic path. Downward direction means negative velocity , After collison ,the velocity becomes positive and velocity decreases Further v1

2=2g(d/2)=gd Therefore v=v1√2 As the direction is reversed and speed is decreased, Hence (a) is the answer

Question 2. The displacement -time graph of a moving particle is shown below.The instantanous velocity of the particle is negative at the point

a. Cb. Dc. Ed. B

Solution(2):. The instantanous velocity is given by the slope of the displacement time graph.Since slope is negative at point B,instantanous velocity is negative at BHence (d) is correct

Question 3.The velocity -time graph of a moving particle is shown below.Total displacement of the particle during the time interval when there is nonzero acceleration and retardation is

a. 60mb. 40mc. 50md. 30m

Solution(3):. Acceleration is given by the slope of the velocity time graphNon zero acceleration happened in the time interval 20 to 40 sec as the slope of the graph is non zero in that time interval

Now displacement in that interval is given by the area enclosed the v-t curve during that interval

So area=(1/2)20*3 +20*1=50mHence (c) is correct Question 4. Figure below shows the displacement -time graph of two particles.Mark the correct statement about their relative velocity

a. It first increases and then decreasesb. It is a non zero constantc. it is zerod. none of the aboveSolution(4):. The instantanous velocity is given by the slope of the displacement time graph.As slope of both the particle displacement time graph is constant.That means there individual velocities are constant.So relative velocity is also constantsso it is a non zero constant as they have different velocityHence (b) is correct

Four position -time graph are shown below.

a.

b.

c.

d.

Question 5 What all graph shows motion with positive velocitiesa. a and c onlyb. all the fourc. b and D onlyd. b only

Solution(5):. The instantanous velocity is given by the slope of the displacement time graph.Since slope is positive in graph a and c.Positive velocity is there in a and c curveHence (a) and (c) are correct

Question 6. What all graph shows motion with negative velocitiesa. a and b only

b. all the fourc. a and c onlyd. c only

Solution(6):. The instantanous velocity is given by the slope of the displacement time graph.Since slope is negative in graph b and d.Negative velocity is there in b and d curveHence (b) and (d) are correct Question 7.which of the folliwing graph correctly represents velocity-time relationships for a particle released from rest to fall under gravity

a.

b.

c.

d.

Solution(7):. The velocity will increase with time so b is the correct answerHence (b) is correct Question 8.The v-x graph of a particle moving along a straight line is shown below.Which of the below graph shows a-x graph

a.

b.

c.

d.

Solution(8):. The equation for the given graph isv=-(v0/x0)x + v0 ---(1)Differentiating both sides we getdv/dx=-v0/x0 ---(2) Nowa=v(dv/dx)ora=(-v0/x0)[-(v0/x0)x + v0 ]ora=mx+cwhere m=v0

2/x0)2

and c=-v02/x0)2

So that means slope is positive and intercept is negativeSo (d) is correct

Question 9.A truck accelerates from rest at the constant rate a for some time after which it decelerates at a constant rate of b to come to the rest.If the total time elapsed is t ,then find out the maximum velocity attains by the trucka. (ab/a+b)tb.(a+b/ab)tc. (a2+b2/ab)td.(a2-b2/ab)t

Solution(9):. Let t1 and t2 be the the time for acceleration and deccleration.

Let v be the maximum velocity attained

Thenv=at1 or t1=v/av=bt2 or t2=v/b

Now t=t1 + t2

or t=v/a + v/b

or v=abt/(a+b)Hence (a) is correct

Question 10.Displacement(y) of the particle is given byy=2t+t2-2t3

The velocity of the particle when acceleration is zero is given bya. 5/2b. 9/4c. 13/6d. 17/8

Solution(10):. given

y=2t+t2-2t3

Velocity is given v=dy/dt=2+2t-6t2

a=dv/dt=2-12t

Now acceleratio is zero2-12t=0t=1/6

Putting this value velocity equationv=2+2/6-6(1/6)2

=2+1/3-1/6=13/6Hence (c) is correct

Question 11.Mark out the correct statementa. Instantaneous Velocity vector is always in the direction of the motionb. Instantaneous acceleration vector is always in the direction of the motionc. Acceleration of the moving particle can change its direction without any change in direction of velocityd. None of the above

Solution(11):. Take the case of uniform circular motion,Instantanous Velocity vector and acceleration vector at any point is tangent and radial to the circle.So it is not along the direction of the circleTake the case the moving car in one direction.If the car accelerated,acceleration is along the direction of velocity.if car driver put a brake then it deacclerates without any change in the direction of the velocity

Hence (c) is correct

Matrix match typeQuestion 12.In a free fall motion from rest,Match column I to column II

column I A) Graph between displacement and timeB) Graph between velocity and timeC) Graph between velocity and displacementD) Graph between KE and displacement

column IIP) ParabolaQ) Straight lineC) CircleD) No appropiate match given

Solution(12):. Equation of motion for a free fall from restx=(1/2)gt2.It is a parabolav=gt it is a straight linev2=2gx it is a parabolaKE=(1/2)mv2=mgx..... it is a straight lineSo the correct answers are A -> PB -> QC -> PD -> Q

13.A body fall from height H.if t1 is time taken for covering first half height andt2 be time taken for second half.Which of these relation is true for t1 and t2

a. t1 > t2

b. t1 < t2

c. t1=t2

d. Depends on the mass of the body

Solution(13):. Let H be the heightthenFirst HalfH/2=(1/2)gt1

2 ----(1)or (1/2)gt1

2=H/2Also v=gt1

Second HalfH/2=vt2+(1/2)gt2

2

or(H/2)=gt1t2+(1/2)gt2

2 or (1/2)gt2

2 =(H/2)-gt1t2 ---(2)

From 1 and 2(1/2)gt2

2=(1/2)gt12 -gt1t2

t22+2t1t2 -t1

2=0

or t2=[-2t1+√(4t1t2 +4t1t2 )]/2or t2=[-2t1+2t1√8]/2 t2=.44t1

so t1 > t2

Hence a is correct

14.A train running at 30m/s is slowed uniformly to a stop in 44 sec.Find the stopping distance?a. 612 mb. 662 mc. 630 md. 605 m

Solution(14):. Here v0=30m/s vf=0 at t=44 secvf=v0+at or 0=30+a(44) or a=-.68m/s2

Now x=v0t+(1/2)at2

or x=662 m

Hence b is correct

15.A nut comes loose from a bolt on the bottom of an elevator as the elavator is moving up the shaft at 3m/s.The nut strikes the bottom of the shaft in 2 sec.How far from the bottom of the shaft was the elevator when nut falls off?a. 13.6 mb. 10 mc. 12.6 md. none of these

Solution(15):. Here the nut initially has the velocity of the elevator ,so choosing upward as positive ,v0=3 m/salso a=-g=-9.8 m/s2

The time taken to hit the bottom is 2 s soy=v0t+(1/2)at2

=-13.6 m

Hence the bottom of the shaft was 13.6 m below elevator when nut fall offHence (a) is correct 16. Let v and a denote the velocity and acceleration respectivly of the particle in one dimension motion.a.the speed of the particle decreases when v.a <0b. the speed of the particle decreases when v.a >0c.the speed of the particle increases when v.a=0d. the speed of the particle decreases when |v|<|a|

Solution(16):. when v.a <0 That mean they are in opposite directionSo when the particle is moving towards origin,acceleration is acting outwards so it is decreasing the speedWhen the particle is moving outwards,acceleration is acting inwards hence it is decreasing the speedSo this option is correct

When v.a > 0That means v and a are in same directionWhen the particle is moving towards origin, acceleration is also acting inwards so increasing the speedWhen the particle is moving outwards,acceleration is also acting outwards hence increasing the speedSo this option is not correct

when v.a=0Then possible cases areAcceleration is zeroVelocity is zero Both are zeroSo speed does not increases in all cases.So this option is not correct

When |v|<|a|Possible casesVelocity is positive ,accleration is positive ----Speed increaseVelocity is negative ,acceleration is negative----Speed increaseVelocity is positive,acceleration is negativeVelocity is negative ,acceleration is positive---Speed decreasesSo this is not correct option

So correct answer is a

17.The displacement of a particle moving in straight line depends on time t asx=at3+bt2+ct+d

which of the following is truea. Initial acceleration depends on b onlyb Initial velocity depends on c onlyc. Initial displacement is dd. Ratio of initial velocity /initial acceleration depends on a and c

Solution(17):. The displacement of a particle moving in straight line depends on time t asx=at3+bt2+ct+d

Velocity (dx/dt)=3at2+2bt+cAcceleration (d2x/dt2)=6at+2b

So Initial displacement(t=0) =dInitial velolcity(t=0) =cInitial acceleration (t=0) =2b

18.A particle located at x=0 at time t=0 starts moving along the positive x-direction with a velocity v that varies asv=√xThe displacement of the particles varies with time as

which of the following is truea. t2

b t3

c. t4

d. t1/2

Solution(18):. Given v=√xordx/dt=√xdx/√x=dt Integrating both sides between the limit (0,x) and (0,t)x=t2/4Hence (a) is correct19.A Train is moving along a straight section of the track with a velocity of 180km/h. The braking deceleration is 2m/s2sAt what distance from a train station should the train driver aply the brake so that train stops at the station a. 800mb 625mc. 700md. none of theseSolution(19):. we are given v0=180km/h=50m/sa=-2m/s2

Now v2=v02+2as

0=(50)2-2*2*x or x=625m hence (b) is correct 20.from the previous question,how long will it take to bring the train to the halt a. 25sb 20sc. 15sd. none of theseSolution(20):. Now v=v0+at0=50-2tor t=25 secHence (a) is correct

Motion in a Plane

Introduction Average velocity Instantaneous velocity Average and instantaneous acceleration Motion with constant acceleration Projectile Motion Uniform circular motion Motion in three dimensions

1. Introduction

In previous chapter we have learned about the motion of any particle along a straight line Straight line motion or rectilinear motion is motion in one dimension.Now in this chapter ,we

will consider both motion in two dimension and three dimension. In two dimensional motion path of the particle is constrained to lie in a fixed plane.Example of

such motion motion are projectile shot from a gun ,motion of moon around the earth,circular motion and many more.

To solve problems of motion in a plane,we need to generalize kinematic language of previous chapter to a more general using vector notations in two and three dimensions.

2.Average velocity

Consider a particle moving along a curved path in x-y plane shown below in the figue Suppose at any time,particle is at the point P and after some time 't' is at point Q where points

P and Q represents the position of particle at two different points.

Position of particle at point P is described by the Position vector r from origin O to P given by r=xi+yjwhere x and y are components of r along x and y axis

As particle moves from P to Q,its displacement would be would be Δr which is equal to the difference in position vectors r and r'.ThusΔr = r'-r = (x'i+y'j)-(xi+yj) = (x'-x)i+(y'-y)j = Δxi+Δyj                                          (1)where Δx=(x'-x) and Δy=(y'-y)

If Δt is the time interval during which the particle moves from point P to Q along the curved path then average velocity(vavg) of particle is the ratio of displacement and corresponding time interval

since vavg=Δr/Δt , the direction of average velocity is same as that of Δr Magnitude of Δr is always the straight line distance from P to Q regardless of any shape of

actual path taken by the particle. Hence average velocity of particle from point P to Q in time interval Δt would be same for any

path taken by the particle.

3.Instantaneous velocity

We already know that instantaneous velocity is the velocity of the particle at any instant of time or at any point of its path.

If we bring point Q more and more closer to point P and then calculate average velocity over such a short displacement and time interval then

where v is known as the instantaneous velocity of the particle. Thus, instantaneous velocity is the limiting value of average velocity as the time interval

aproaches zero. As the point Q aproaches P, direction of vector Δr changes and aproaches to the direction of

the tangent to the path at point P. So instantaneous vector at any point is tangent to the path at that point.

Figure below shows the direction of instantaneous velocity at point P.

Thus, direction of instantaneous velocity v at any point is always tangent to the path of particle at that point.

Like average velocity we can also express instantaneous velocity in component form

where vx and vy are x and y components of instantaneous velocity. Magnitude of instantaneous velocity is

|v|=√[(vx)2+(vy)2]and angle θ which velocity vector makes with x-axis istanθ=vx/vy

Expression for instantaneous velocity is

Thus, if expression for the co-ordinates x and y are known as function of time then we can use equations derived above to find x and y components of velocity.

4. Average and instantaneous acceleration

Suppose a particle moves from point P to point Q in x-y plane as shown below in the figure

Suppose v1 is the velocity of the particle at point P and v2 is the velocity of particle at point Q Average acceleration is the change in velocity of particle from v1 to v2 in time interval Δt as

particle moves from point P to Q. Thus average acceleration is

Average accelaration is the vector quantity having direction same as that of Δv. Again if point Q aproaches point P, then limiting value of average acceleration as time

aproaches zero defines instantaneous acceleration or simply the acceleration of particle at that point. Ths, instantaneous acceleration is

Figure below shows instantaneous acceleration a at point P.

Instantaneous acceleration does not have same direction as that of velocity vector instead it must lie on the concave side of the curved surface.

Thus velocity and acceleration vectors may have any angle between 0 to 180 degree between them.

5. Motion with constant acceleration

Motion in two dimension with constant acceleration we we know is the motion in which velocity changes at a constant rate i.e, acceleration remains constant throughout the motion

We should set up the kinematic equation of motion for particle moving with constant acceleration in two dimensions.

Equation's for position and velocity vector can be found generalizing the equation for position and velocity derived earliar while studying motion in one dimensionThus velocity is given by equationv=v0+at                                          (8)wherev is velocity vectorv0 is Intial velocity vectora is Instantanous acceleration vector

Similary position is given by the equationr-r0=v0t+(1/2)at2                                          (9)

where r0 is Intial position vectori,er0=x0i+y0jand average velocity is given by the equationvav=(1/2)(v+v0)                                          (10)

Since we have assumed particle to be moving in x-y plane,the x and y components of equation (8) and (9) arevx=vx0+axt                                          (11a)x-x0=v0xt+(1/2)axt2                                          (11b) and vy=vy0+ayt                                           (12a)y-y0=v0yt+(1/2)ayt2                                           (12b)

from above equation 11 and 12 ,we can see that for particle moving in (x-y) plane although plane of motion can be treated as two seperate and simultanous 1-D motion with constant acceleration

Similar result also hold true for motion in a three dimension plane (x-y-z)

6. Projectile Motion

Projectile motion is a special case of motion in two dimension when acceleration of particle is constant in both magnitude and direction

An object is referred as projectile when it is given an intial velocity which subsequently follows a path determined by gravitational forces acting on it.For example bullet fired from the rifle,a javalin thrown by the athelite etc

Path followed by a projectile is called its trajectory In this section we will study the motion of projectile near the earth surface neglecting the air

resistance. Acceleration acting on a projectile is constant which is acceleration due to gravity (g=9.81

m/s) directed along vertically downward direction. we shall treat the projectile motion ina cartesian co-ordinates system taking y axis in vertically

upwards direction and x axis alomg horizontal directions Now x and y components of acceleration of projectile is

ax=0 and ay=-gSince acceleration in horizontal direction is zero,this shows that horizontal component of velocity is constant and verical motion is simply a case of motion with constant acceleration.

Suppose at time t=0 object is at origin of co-ordinate system and velocity components v0x and v0y.From above components of acceleration are ax=0 and ay=-g.From equation 11 and 12 in the previous section components of position and velocity arex=v0xt                                          (14a)vx=v0x                                           (14b)

and vy=v0y-gt                                          (15b)y=v0y-(1/2)gt2                                          (15a)

Figure below shows motion of an object projected with velocity v0 at an angle θ0

In terms of initial velocity v0 and angle θ0 components of initial velocity arev0x=v0cosθ0                                          (16a)v0y=v0sinθ0                                           (16b)

Using these relations in equation 14 and 15 we findx=(v0cosθ0)t                                          (17a)y=(v0sinθ0)t-(1/2)gt2                                           (17b)vx=v0cosθ0                                           (17c)vy=v0sinθ0-gt                                           (17d)

Above equations describe the position and velocity of projectile as shown in fig 5 at any time t.

(A)Equation of Path of projectile(Trajectory)

From equation 17at=x/v0cosθ0 now putting this value of t in equation 17b,we findy=(tanθ0 )x-[g/2(v0cosθ0)2]x2                                           (18)

In equation (18),quantities θ0,g and v0 are all constants and equation (18) can be compared with the equation y=ax-bx2

where a and b are constants

This equation y=ax-bx2 is the equation of the parabola.From this we conclude that path of the projectile is a parabola as shown in figure 5

B) Time of Maximum height

At point of maximum height vy=0.Thus from equation (17d)vy=v0sinθ0-gt 0=v0sinθ0-gt -or tm=v0sinθ0/g                                          (19)

Time of flight of projectile which is the total time during which the projectile is in flight can be obtained by putting y=0 because when projectile reaches ground ,verical distance travelled is zero.This from equation (17b)

tf=2(v0sinθ0)/g                                          (20)ortf=2tm

Maximum height reached by the projectile can be calculated by substituting t=tm in equation 17by=Hm=(v0sinθ0)(v0sinθ0/g)-(g/2)(v0sinθ0/g)2

orHm=v0

2sin2θ0/2g                                          (21)

(C) Horizontal Range of Projectile

Since acceleration g acting on the projectile is acting vertically ,so it has no component in horizontal direction.

So, projectile moves in horizontal direction with a constant velocity v0cosθ0. So range R isR=OA=velocity x time of flight

Maximum range is obtained when sin2θ0=1 or θ0=450. Thus when θ0=450 maximum range achieved for a given initial velocity is (v0)2/g.

7. Uniform circular motion

When an object moves in a circular path at a constant speed then motion of the object is called uniform circular motion.

In our every day life ,we came across many examples of circular motion for example cars going round the circular track and many more .Also earth and other planets revolve around the sun in a roughly circular orbits

Here in this section we will mainly consider the circular motion with constant speed if the speed of motion is constant for a particle moving in a circular motion still the particles

accelerates becuase of costantly changing direction of the velocity. Here in circular motion ,we use angular velocity in place of velocity we used while studying

linear motion

(A) Angular velocity

Consider an object moving in a circle with uniform velocity v as shown below in the figure

The velocity v at any point of the motion is tangential to the circle at that point.Let the particle moves from point A to point Balong the circumference of the circle .The distance along the circumference from A to B iss=Rθ                                          (23)Where R is the radius of the circle and θ is the angle moved in radian's

Magnitude of velocity isv=ds/dt=Rdθ/dt                                          (24)Since radius of the circle remains constant quantity,ω=dθ/dt                                          (25)is called the angular velocity defined as the rate of change of angle swept by radius with time.

Angular velolcity is expressed in radians per second (rads-1) From equation 24 and 25,we find the following

v=ωR                                          (26) Thus for a particle moving ain circular motion ,velocity is directly proportional to radius for a

given angular velocity For uniform circular motion i.e, for motion with constant angular velocity the motion would be

periodic which means particle passes through each point of circle at equal intervals of time Time period of motion is given by

T=2π/ω                                          (27)Since 2π radians is the angle θ in one revolution

If angular velocity ω is constant then integrating equation (25) with in limits θ0 to θ,we find

where θ0 is the angular position at time t0 and θ is the angular position at time t .The above equation is similar to rectilinear motion result x-x0=v(t-t0)

(B) Angular acceleration

Angular acceleration is defined as the rate of change of angular velocity moving in circular motion with time.Thusα=dω/dt=d2θ/dt2                                          (29)Unit of angular acceleration is rads-2

For motion with constant angular acceleration

or ω=ω0+α(t-t0)                                          (30)where ω0 is the angular velocity at time t0

Again sinceω=dθ/dt or dθ=ωdt then from equation 30

If in the begining t0=0 and θ0=0 the angular position at any time t is given byθ=ωt+(1/2)αt2

This result is of the form similar to what we find in case of uniformly accelerated motion while studying rectilinear motion

8. Motion in three dimensions

We have already studied physical quantities like displacement ,velocity,acceleration etc in one and two dimension

In this topic ,we will generalize our previous knowlegde of motion in 1 and 2 -dimension to three dimension's

As we have used vectors to represent motion in a plane,we can freely use vectors and its properties in 3-dimension as we have done in case of motion in a plane

In three dimensions ,we have three units vectors i ,j and k associated with each co-ordinate axis of cartesian co-ordinates system shown below in the figure

Consider a particle moving in 3-D space .Let P be its position at any point t.Position vector of this particle at point P would ber=xi+yj+zkWhere x,y and z are co-ordinates of point P

Similarly velocity and acceleration vectors of particle moving in 3-D space arev=vxi+vyj+vzk where vx=dx/dt,vy=dy/dt and vz=dz/dtanda=vxi+ayj+azkwhere ax=dvx/dt,ay=dvy/dt and az=dvz/dt

All the relations we have derived incase of motion in plane are valid for 3-D motion with one added co-ordinate

Test Series-7

Instructions

Your test contains 12 multiple choice questions with only one answer This is 20 min test.Please make sure you complete it in stipulated time You can Finish this test any time using 'Submit' button. Once finished you get a chance to review all question with correct answers and their

solutions.

1.A ball is projected upward at a certain angle with the horizontal .which of the following statement is incorrect. At highest point

velocity of the projectile is not zero

acceleration of the projectile is zero

velocity of the projectile is along the horizontal direction

None of these

2.what is wrong for a body having uniform circular motion?

Speed of the body is constant

Acceleration is directed towards the centre

Velocity and Acceleration vector are having an angle 45

none of the above

3.A hunter aims his guns and fires a bullet directly at a monkey on a tree.At the instant bullet leaves the gun,the monkey drops. The bullet

Misses to hit the monkey

Hits the monkey

cannot be said

None of these

4.Two bullets A and B are fired horizontally with speed v and 2v respectively.which of the following is true

Both will reach the ground in Same time

Bullet with speed 2v will cover more horizontal distance on the groud

B will reach the ground in less time then A

A will reach the ground in less time then B

5.A football is kicked into the air at an angle of 30 degrees with the horizontal. At the very top of the ball's path, its velocity is

entirely horizontal

both vertical and horizontal

not enough information given to know.

entirely vertical

6. Which of the following statements are true

The time that a projectile is in the air is dependent upon the vertical component of the initial velocity.

The vertical acceleration of a projectile is 0 m/s/s when it is at the peak of its trajectory.

Horizontal velocity changes in the projectile motion

none of the above

7. which of the following is incorrect expression for three dimensional motion

a=dv/dt=(d2x/dt2)i+(d2y/dt2)j+(d2z/dt2)k

a=vdv/dr

v=dr/dt=(dx/dt)i + (dy/dt)j+(dz/dt)k

None of these

8. Four statment are made about the projectile motionSTATEMENT -1: Horizontal velocity remains same through out the motionSTATEMENT -2: The trajactory is a parabolaSTATEMENT -3: Acceleration of the projectile is vertically upwardsSTATEMENT -4: Momentum remains constants through out the motionwhich one of the following is correct

All the statement are correct

Statement 1,2 and 4 are correct

Statement 1,2 and 3 are correct

Statement 2,3 and 4 are correct

9.which of the following is not a projectile motion

A football thrown at an angle

A bomb falling from the aircraft

A parchutor falling from aeroplane

None of these

10.Which of the following is correct expression for Projectile

Range=2uxuy/g

Magnitude of velocity at height h=(u2+2gh)1/2

T=2ux/g

None of these

11. The trajectory of a projectile in a vertical plane is given byy=2x-4x2

where x and y are the horizontal and vertical distanceFollowing statement are made about this motionSTATEMENT A: The angle of projection is given by tan-12STATEMENT B:The Range of the Projectile is .5 units

A and B both are correct

Only A is correct

Only B is correct

Both A and B are wrong

12. A parachuter falls from an aeroplane moving with uniform horizontal velocity.Which of the following is false

when he land on the surface,he will be just below the aeroplane

A person sitting in the aeroplane will see the parachuter going vertically downward

when he land on the surface,he would left behind the aeroplane

none of these