1.2 - Equations of Motion
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Transcript of 1.2 - Equations of Motion
AccelerationWhen an unbalanced force acts on an object it accelerates.An unbalanced force acting on a stationary object will make it move. An unbalanced force can make a moving object get faster or slower or change direction.
Acceleration is the change in velocity per unit time.
The acceleration or deceleration of an object is calculated using:
tu-va
time (s)
acceleration(ms-2)
initial velocity (ms-
1)
final velocity (ms-1)
Measuring Acceleration
DiagramElectronic
Timer
Light GateLength of Card
Trolley
Start Line
Finish Line
Stopwatch
MethodA trolley is fitted with a length of card. The length of card is measured using a ruler.The trolley is released from rest, so the initial velocity is 0 ms-
1.The time taken for the trolley to travel from the start to the finish line is measured using a stopwatch.An electronic timer measures the time taken for the length of card to pass through the light beam. The timer starts when the card breaks the light beam and stops when the card leaves the light beam.
The final velocity is calculated using:
These measurements are used to calculate acceleration in the relationship shown:
timecard of lengthv
tuva
Results
-1ms 0u
1-ms timecard of lengthv
s t
?a
Motion GraphsThere are three types of motion graph we will study:
1. A velocity - time graph.2. A displacement – time
graph.3. An acceleration – time
graph.
displacement-time
velocity-time acceleration-time
Constant Velocity
Constant Acceleration
t
s
t
v
t
a
t
s
t
v
t
a
Constant Deceleration
ta
t
v
t
s
Drawing GraphsEach velocity – time graph has a corresponding acceleration – time graph and displacement- time graph.
ExampleThe following velocity – time graph describes a journey.
time / s
velocity / ms-1
3 10 12
12
0
Draw the corresponding acceleration – time graph.
tuva
3012
2ms 4a
tuva
71212
2ms 0a
tuva
2120
2ms -6a
0-3 seconds 3-10 seconds
10-12 seconds
time / s
acceleration / ms-2
3
4
010 12
-6
Q1. Copy out the following velocity - time graph and underneath it draw the corresponding acceleration – time graph.
time / s
velocity / ms-1
4 13 16
20
0 9
24
tuva
4020
2ms 5a
2ms 0a t
uva
42024
2ms 1a
0-4 seconds 4-9 seconds 9-13 seconds
tuva
3240
2ms 8a
13-16 seconds
acceleration / ms-2
time / s4
5
09 13
- 8
16
1
Q2. Using the following acceleration – time graph of an object starting from rest, draw the corresponding velocity – time graph.
acceleration / ms-2
0
5
-24 8 1
8time / s
atuv
450 -1ms 20v
atuv
102-20
2020 -1ms 0v
velocity / ms-1
0
20
4 8 18
time / s
Q3. Copy out the following velocity – time graph and underneath it draw the corresponding acceleration – time graph (after appropriate calculations).
velocity / ms-1
time / s
3
15
0
-9
3 7 9 14 2418
tuva
3315
2ms 4a
0-3 seconds 3-7 seconds2ms 0a
tuva
2153
2ms -6a
7-9 seconds
9-14 seconds
2ms 0a t
uva
439-
2ms -3a
14-18 seconds
tuva
6
9-0
2ms 1.5a
18-24 seconds
acceleration / ms-2
0
4
-6
3 7 9 14 2418 time / s
Motion and DirectionIn the velocity-time graphs so far, motion has been in only one direction.A velocity time graph however can show two different directions; one direction is positive, while the other is negative.
Example 1A car travelling along a straight section of road has the following velocity-time graph.
(a) Calculate the distance travelled by the car.(b) Calculate the displacement of the car.
forward motion
backward motion
(a) Distance travelled is the area under velocity-time graph.
2010021d1 10502
1d2
10100d3 105021d4
m 1,000
m 1,000 m 250
m 250
(b) Displacement is the forwards motion, less the backward motion.
2501,000250 - 1,000 ntdisplaceme
m 500
m 2,500dtot
Worksheet – Graphs of Motion
Q1 – Q9
Ball Thrown Into AirA ball is thrown directly upwards into the air.It rises into the air and falls back down to the thrower.The velocity – time graph and corresponding acceleration – time graph are shown.
velocity / ms-1
time / s
acceleration / ms-2
time / s
-9.8
falling
Bouncing Ball (No Energy Loss)
A ball is dropped from a height to the ground.The ball bounces twice with no energy loss and is then caught.The velocity - time and acceleration - time graphs are as follows:
time / s
velocity / ms-1
downwards (falling): - ve upwards (rising): +
ve
rising
falling
rising
hits ground
max height
time / s
acceleration / ms-2
-9.8
rising
TASKRe-draw the velocity-time graph for the bouncing ball, taking downward motion as positive, and upward motion as negative.
fallin
g
time / s
velocity / ms-1
downwards (falling): + ve upwards (rising): -
ve
hits ground
max height
rising
fallin
g
Bouncing Ball (With Energy Loss)
A ball is dropped from a height to the ground.Kinetic energy is lost with each bounce.The velocity - time and acceleration - time graphs are as follows:
time / s
velocity / ms-1
time / s
acceleration / ms-2
-9.8
TASKRe-draw the velocity-time graph for the bouncing ball, taking downward motion as positive, and upward motion as negative.
Worksheet – More Graphs of Motion
Q1 – Q7
Equations of Motion
1st Equation of Motion
tuva
t auv
t auv
There are three equations of motion.You must be able to use and derive these three equations of motion.
2nd Equation of MotionTo derive the second equation of motion, the velocity-time graph shown is used as a starting point.
time/s
velocity/ms-
1
u
v
t
} v - u
Displacement (s) is the area under a velocity time graph.
area areas
u-v t21 ut
at t21 ut s
but from equation (1):
atuv
atu-v
so we can rewrite as:
2at21 ut s
atuv
2atuv2
atuatuv2 2222 ta2uatuv
taking a common factor of 2a gives
222 at2
1ut2auv
and since s = ut + ½at2
2asuv 22
3rd Equation of MotionThis equation links final velocity (v) and displacement (s).
DirectionWhen using the equations of motion, it is essential that direction is considered.In these examples, upward motion is taken as positive, so any downward motion is taken to be negative.
ExampleA helicopter is travelling upwards with a velocity of 25 ms-1. A package is released and hits the ground 14 s later. *
*
*
released
stationary ( 0 ms-1
)
hits ground
Path of Package
This example has motion in two directions.It is necessary to distinguish between the two directions.
Choose the upward direction as positive!
(a) How long will it take the package to reach its maximum height? (2)
?t 1ms 25 u
1ms 0v 2ms 9.8a
atuv
t9.8-250
25t 9.8 s 2.55t
(b) How high as it climbed since being released?(2)
?s 1ms 25 u
s 2.55t 2ms 9.8 a
2atuts21
22.559.8-2.5525 21
31.8663.75
m 31.9s
(c) Calculate the velocity of the package just before it hits the ground.(2)
?v1ms 25 u
s 14 t 2ms 9.8a
atuv 149.8-25
137.2-251ms 112.2v
The negative indicates travelling downwards.
(d) How high above the ground is the helicopter when the package is released? (2)
?s 1ms 25 u
s 14t 2ms -9.8a
2atuts21
2149.8-211425
960.4350
m 610.4s
So the helicopter is 610.4 m above the ground.
Worksheet – Equations of Motion
Q1 – Q15
Acceleration Due To Gravity
Diagram
Ball Bearing
Trap Door
Timer
MethodThe ball bearing is released from rest, so initial velocity (u) is 0 ms-1.The displacement (s) of the ball bearing is the distance between the release point and trap door, and is measured using a metre stick.An electronic timer measures the length of time taken for the ball bearing to reach the trap door.
-1ms 0um s
s t
?a
2atuts21
Calculation
Projectiles (Half Trajectory)
An object projected sideways through the air will follow a curved trajectory.
horizontal motion (steady speed)
vertical motion
The horizontal and vertical motions should be treated separately.Time is the only quantity common to both.
accelerates downwards at -9.8 ms-
2
tDV H
H
vh
vv
This is an example of a ‘half-trajectory.’
GREEN – actual motionRED – vertical motionBLUE – horizontal motion
At any point in its trajectory, the velocity of a projectile has two components.
• one vertical, VV
• the other horizontal, VH
The resultant velocity is found drawing a vector diagram and add the vectors together, TIP to TAIL.
Vector Diagram
horizontal velocity
vertical velocity
resultant/actual
velocity
30 ms-1
ExampleA ball is kicked horizontally off an embankment, with a velocity of 30 ms-1.It lands 24 m from the base of the embankment.
(a) Calculate how long the ball was in flight.
24 m
tDV H
H
t2430
3024t
s 0.8t common to
horizontal and vertical motions
Horizontal Vertical-1ms 30u
m 24s
-2ms -9.8a
(b) Calculate the horizontal velocity just before hitting the ground.
s 0.8t
s 0.8t
-2ms 0a
-1ms 0u
travels horizontally at steady speed
– no acceleration horizontally
not initially falling down, so speed of
zero in vertical direction
acted upon by gravity
Horizontal
atuv
0.8030 -1ms 30v
(c) Calculate the vertical velocity just before hitting the ground.
Vertical
atuv
0.89.8-0 -1ms -7.84v
(d) How high is the embankment?
Vertical-2ms -9.8a
s 0.8t -1ms 0u
-1ms -7.84v
means 7.84 ms-1 downwards
2at21uts
20.89.8-210.80
m 3.14s
so height of the embankment is 3.14 m
means ball fell through distance of
3.14 m
(e) Calculate the resultant velocity of the ball, just before hitting the ground.
30 ms-
1
-7.8 ms-1
velocity
θ
SizeBy Pythagoras:
222 cba
222 7.84-30velocity resultant 61.5900
961.5velocity resultant -1ms 31
Direction
hypadjθ cos
3130θ cos
0.97cosθ 1
14.6θ
horizon below 14.6 of angle at ms 31velocity resultant -1
30 ms-
1
-7.8 ms-1
velocity
θ
Q1. A ball is kicked off a cliff with a horizontal speed of 16 ms-1.
The ball hits the ground 2.2 s later.
(a) Calculate the height of the cliff.(b) Calculate the distance between the foot of
the cliff and where the ball lands.(c) Calculate the vertical component of the
balls velocity just before it hits the ground.
(d) Calculate the balls velocity as it hits the ground.
23.7 m
35.2 m
21.6 ms-1
26.9 ms-1 at angle of 53.5° below
horizon
You may want to draw a diagram to help you get started !!!
Q2. A ball is kicked off a cliff with a horizontal speed of 22 ms-1. the ball hits the ground 1.5 s later.
(a) Calculate the height of the cliff.(b) Calculate the horizontal distance from the
foot of the cliff, to where the ball lands.(c) Calculate the vertical component of the balls
velocity as it hits the ground.(d) Calculate the balls actual velocity as it hits
the ground.
11 m
14.7 ms-1
26.5 ms-1 at angle of 34° below
horizon
33 m
You may want to draw a diagram to help you get started !!!
Does Projectile Theory Work?
Diagramball-bearing
h
d
sand
MeasurementsHorizontal Velocity measure distance ball-bearing travels
along desk and divide by time taken
Vertical Displacement measure height of desk from floor
CalculationCalculate the time of flight.
m s
-2ms -9.8a-1ms 0u
?t
2at21uts
Vertical
Now calculate the horizontal displacement.
-1H ms v
Horizontal
s t
?s
tvs H
Experimentally
The horizontal displacement was measured experimentally using a metre stick to be m.
Projectiles (Full Trajectory)
A projectile does not need to be an object falling, but could be an object fired at angle to the horizontal.
The subsequent motion would bemax height
θ
If air resistance is ignored, the trajectory has an axis of symmetry about the mid point (maximum height).So the time taken to reach the maximum height is the same as the time taken to fall back to the ground.
Various calculations can be made, but firstly, the initial velocity must be split into its horizontal and vertical components.
Horizontala = 0 ms-2
Verticala = -9.8 ms-2
Example 1A golf ball is hit off the tee at 48 ms-1 at angle of 20° to the horizontal.
Calculate the horizontal and vertical components of the initial velocity.
Horizontal
20°
48 ms-1
hypadjθ cos
48v20 cos H
cos2048vH -1
H ms 45.1v
Vertical
hypoppθ sin
48v20 sin V
20 sin48vV -1
V ms 16.4v
48 ms-1
20°VH
VV
Example 2An arrow is projected into the air with a velocity of 38 ms-1 at an angle of 25° to the horizontal.
38 ms-1
250
(a) Calculate the horizontal and vertical components of the initial velocity.
Horizontal
hypadjθ cos
38v25 cos H
25 cos38vH -1
H ms 34.4v
Vertical
hypadjθ sin
38v25 sin V
25 sin38vV -1
V ms 16.1v
38 ms-1
25°VH
VV
(b) Calculate the arrow’s maximum height.
Vertical-2ms -9.8a-1ms 16.1u
-1ms 0v
?s
2asuv 22
s-9.8216.10 22
s 19.6259.210
259.21s 19.6
m 13.2s
(c) Calculate the time taken for the arrow to reach its maximum height.
Vertical-2ms -9.8a-1ms 16.1u
-1ms 0vm 13.2s
?t
atuv
t9.8-16.10 16.1t 9.8
s 1.64t
(d) Calculate the total time of the arrows flight.
down time up timetime total
1.64 1.64
s 3.28
(e) Calculate the horizontal distance travelled by the arrow until impact with the ground.
Horizontal-1ms 34.4u
-2ms 0as 3.28t
?s
2at21uts
23.280213.2834.4
112.8ms
(f) Calculate the arrow’s velocity 0.5 s after being fired.
Firstly calculate the vertical component of velocity (horizontal component is constant, since a = 0 ms-2)
Vertical-2ms -9.8a-1ms 16.1u
s 0.5t ?v
atuv
0.59.8-16.1
4.9-16.11ms 11.2v
Now calculate the actual velocity after combining the vertical and horizontal components of the velocity after 0.5 s.
v
θ34.4 ms-
1
11.2 ms-1
SizeBy Pythagoras: 222 cba
222 11.234.4velocity 125.441,183.36
1,308.8velocity-1ms 36.2
Direction
adjoppθ tan
34.411.2θ tan
0.326tanθ 1
18θ
Velocity of the arrow after 0.5 s is:
36.2 ms-1 at angle of 18° above the horizon
Q1. A shell is fired from a gun with a velocity of 72 ms-1 at an angle of 60° to the horizontal.
(a) Calculate the horizontal and vertical components of the initial velocity.
(b) Calculate the maximum height reached.(c) Calculate the time taken for the shell to
reach it’s maximum height.(d) Calculate the total time of flight.(e) Calculate the horizontal range of the shell.(f) Calculate the shells velocity after 2.3 s.
VH = 36 ms-1
VV = 62.4 ms-1
199 m
6.4 s
12.8 s
458 m
53.7 ms-1 at angle of 48° above the horizon
Q2. An arrow is fired with a velocity of 50 ms-1 at an angle of 30° to the ground.
(a) Calculate the time taken for the arrow to reach its maximum height.
(b) Calculate the maximum height reached by the arrow.
(c) Calculate the time the arrow is in flight.(d) Calculate how far away from the firing point
the arrow will land.(e) Calculate the actual velocity of the arrow 1s
after it is fired.
2.55 s
31.89 m
5.1 s
5.1 s
45.89 ms-1 at angle of 19.3° above horizon