1st year college physics questions

8/10/2019 1st year college physics questions

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velocity (m/s)

Time (s)

5

10

15

5 10 15

15 20 25 30

-5

PROBLEM 2.1. ANSWER

THE FOLLOWING

QUESTIONS ABOUT A

PARTICLE WHOSEVELOCITY IS PLOTTED

VERSUS TIME TO LEFT

(ORANGE LINE)

(a) Use the velocity graph above to plot quantitatively a graph of

acceleration versus t ime, with correct scales on both axes.

acceleration (ms-2)

Time (s)

2

5 10 15

15 20 25 30

-2

2 1

2 1

5 5Motion from t=0 to t=10s: accln is 0 ( constant)10 0

v va t t

= = = =

215 5Motion from t=10s to t=15s: accln is 2 / ( constant)15 10

f i

f i

v va m s

t t

= = = =

2( 5) 15 20Motion from t=15s to t=25s: accln is 2 / ( constant )25 15 10

f i

f i

v va m s

t t

= = = = =

( ) 2( 5) 5 0Motion from t=25s to t=30s: accln is 0 / ( constant )30 25 5

f i

f i

v va m st t

= = = = =


2/5

(b) Find the displacement (change of position) between t = 10 sec and t = 15 sec.

25

velocity (m/s)

Time (s)

5

10

15

5 10 15

15 20 30

-5

WORKING TO GET

AREA BETWEEN

VELOCITY VS T

GRAPH AND TIME

AXIS

Displacement f ix x

area between v t graph and t axis

= =

+

= (base)(ht) +(1/2)(base)(ht)

= 5X5 + (1/2)(5)(10)= 50 m

(c) Find the displacement between t = 20 sec and t = 25 sec. +

= (1/2)(base)(ht) +(1/2)(base)(ht)

= (1/2)(2.5)(5) + (1/2)(2.5)(-5)= 6.25 6.25 = 0 m

Note area below axis is negative


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Answer 2. The shortest possible runway has a length equal to the distance s

travelled by the plane while coming to rest. (But this would leave the pilot withno room for error the wheels would have to touch down exactly at the near end

of the runway!)

This is a problem of constant acceleration so we use one of the Eqs from slide #

28 of the Mechanics A Lecture Slides.

Given: Initial velocity v0 = 220 kph = 220/3.6 m/s = 61.11 m/s (converted to SIunits)

Time t = 29 sec

Final velocity = 0 (because the plane comes to rest)

Wanted: displacement (distance) s = ?

The acceleration a is missing neither given nor wanted.

Therefore we use Eq 3 from Lecture Slide 29, , which also has a missing:

886 metre.

This is the shortest possible length of a runway

Since only 2 signi ficant figures were given for the time, best quote 8.9X102 m

Problem 2

A jetliner touches down at 220 km/h, reverses its engines to provide braking, andcomes to a halt 29 sec later. What is the shortest runway on which this aircraft

can land, assuming constant deceleration starting from touchdown?

0 61.11 0 292 2

v vs t

+ += = =


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Problem 3

Launched from the ground, a rocket accelerates vertically upwards at 4.6 ms-2. It

passes through a band of clouds 5.3 km thick, extending upwards from an altitude

of 1.9 km. How long is it in the clouds?

y=0

y=1.9 km

y=7.2 km

In this problem we are dealing with data involving three

spatial locations A(y=0), B(y=1.9 km) and C(y = 1.9+5.3 km = 7.2 km). A single equation-of-constant-

acceleration can only deal with two spatial locations (an

initial point and a final point). Therefore we will need TWO

applications of constant-acceleration equation(s).

There are (at least) two different strategies that will solve

this problem.Strategy 1.

(a) Apply eq of const accel to find time tAB from ground to

cloud bottom.

(b) Apply eq of const accel to find time tAC from ground to

cloud top.

(c) Time in cloud is tBC= tAC - tABStrategy 2.

(a) Solve motion from A to B to get velocity at B, vB.

(b) Use vB as the initial velocity in motion B to C, hence

getting tBC.

A

B

C

groun

dd

cloud

Solution 3


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Solution of Problem 3 via Strategy 1

(a) For const.-acceln. motion A to B (choosing upwards as positive):v0 = 0, a = +4.6 ms

-2, s = 1.9 km = 1.9103 m. t = ?

Final velocity v is missing so use s = v0t + at2

1.9103 = 0 + (4.6)t2

t2 = (1.9103)(2)/(4.6) = 826.09

t = (826.09) = 28.742 sec = tAB(b) For const.-acceln. motion A to C (choosing upwards as positive):

v0 = 0, a = +4.6 ms-2, s = 7.2 km = 7.2103 m. t = ?

By same working as above tAC = [(7.2103)(2)/(4.6)] = 55.95 sec.

(c) Then time in cloud = tAC tAB = 55.95 28.742 = 27.208 sec

Solution of Problem 3 via Strategy 2.(a) for const acel motion from A to B (upwards positive) : v0 = 0, a = +4.6 ms

-2, s = +1.9103 m; v = ?

Here t is missing, so we use v2 = v02 +2as

= 02 + 2(4.6)1.9103 = 17480

v = 17480 = 132.211

(b) for const acel motion from B to C (upwards positive): v0 = 132.211, a= +4.6 ms-2, s = +5.3103 m, t = ?

Here v is missing so we use s = v0t + at2(a)t2 + (v0)t + (s ) = 0

Solution of quadratic for t f rom quadratic formula:

= 27.208 sec ( + solution used: - solution unphysical)

Ans: by either Strategy, solut ion is Rocket Spends 27 sec in the cloud layer(Quot ing more than 2 decimal places is not just ified by precision of input data)

1st year college physics questions

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