# Free Motion 01 Practice Test - 9th Grade

A coin is thrown in a vertically upward direction with a velocity of 5 ms1. If the acceleration of the coin during its motion is 10 ms2 in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?

A. s=1.5 m, t=0.25 s
B. s=0.25 m, t=1.5 s
C. s=1.25 m, t=0.75 s
D. s=1.25 m, t=0.5 s

#### SOLUTION

Solution : D

Given, initial velocity, u=5 ms1, acceleration, a=10 ms2.

Let 'v' be the final velocity and 's' be the height attained.
At the highest point of motion, v=0

Using third equation of motion, v2=u2+2as,
0=52+[2×(10)×s]
s=1.25 m
Hence the height attained by the coin is 1.25 m.

Let the time taken be t.
Using the first equation of motion,
v=u+at
0=510t
t=0.5 s
Time taken by the coin is 0.5 s

A particle moves 3 m north, then 4 m east and finally 6 m south. What is the distance travelled and the displacement of the particle?

A. The distance travelled is 13 m.
B. The distance travelled is 12 m.
C. The displacement is 13 m.
D. The displacement is 5 m.

#### SOLUTION

Solution : A and D Let the particle start from A and finally reach B as shown. Distance is the length of the actual path taken by it.
In this case,
distance = 3 m + 4 m + 6 m = 13 m
Displacement is the shortest distance between the initial and final position.
Here, displacement = AB
Using Pythagoras theorem,
AB =(BC2+AC2)
=(32+42) m=5 m.

An express train going towards Delhi is travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of 0.5m s2 . Find how far the train will go before it is brought to rest.

A.

600 m

B.

725 m

C.

625 m

D.

500 m

#### SOLUTION

Solution : C

Given,
initial velocity, u=90 kmph = 90×518 =25 ms1
acceleration, a= 0.5 ms2
final velocity, v=0  (brakes are applied)
Let 's' be the distance travelled.
From the third equation of motion,
v2=u2+2as

0=252+(2×0.5×s)

s=625 m

Hence the distance covered is 625 m.

In the given figure, the velocity of the body at A is zero. A.

True

B.

False

#### SOLUTION

Solution : A The slope of a displacement-time graph gives us the velocity. The slope at any point in a curve can be found by drawing a tangent on it. In the figure, a tangent has been drawn at point A.
Slope=Change in y-coordinateChange in x-coordinate
At point A, clearly, there is no change in the y-coordinate. Hence, the slope is zero. This implies that the velocity at point A is zero.

A 60 m long train moving on a straight level track passes a pole in 5 s. Find the speed of the train and the time it will take to cross a 540 m long bridge.

A. 13 ms1, 40 s
B. 25 ms1, 30 s
C. 12 ms1, 50 s
D. 15 ms1, 65 s

#### SOLUTION

Solution : C

Given:
Length of the train, D=60 m,
Time taken to cross the pole, T=5 s,
Length of the bridge =540 m

Speed of the train, s=DT

Speed of the train, s=605=12 ms1

Now, distance to be covered to cross a 540 m long bridge is sum of Length of the train and Length of the bridge.
d=60+540=600 m

Time taken=ds

Time taken=60012=50 s

Speed of the train is 12 ms1 and the time it will take to cross the bridge is 50 s.

An auto travels the first half time with a uniform speed u and the next half time with a uniform speed v. Find its average speed.

A. u+v2
B. uv2
C. 2uv
D. uv

#### SOLUTION

Solution : A Given, speed of the auto during the first half of  the journey is u and the next half of the journey is v
Let total time taken for the journey be T
Time taken for the first half of the journey be T2 and the next half of the journey be  T2
Let the average speed be Vav
We know, speed=Distancetime
Distance=speed×time

During the first half of the journey, distance, d1=u×T2
During the second half of the journey, distance, d2=v×T2
Total distance  =d=d1+d2
=u×T2+v×T2
d=T2(u+v)

Average speed,Vav =  Total distanceTotal time taken
=dT
=T2(u+v)T
=u+v2

An artificial satellite is moving in a circular orbit of radius 4200 km around the earth. What is its speed if it takes one day to complete one revolution?

A. 4099 kmph
B. 3099 kmph
C. 1099 kmph
D. 2099 kmph

#### SOLUTION

Solution : C

Given, the radius of the orbit, R=4200 km
The time taken to revolve around the Earth, T=24 hr (1 day)
The speed of the satellite, v=Circumference of the orbitTime taken
=2πRT
=2×3.14×4200 km24 hr
=1099 km hr1
=1099 kmph

Motion of a satellite in a circular orbit is an example of

A. non-uniform circular motion
B. uniform circular motion
C. uniform linear motion
D. non-uniform linear motion

#### SOLUTION

Solution : B

The motion of a satellite in a circular orbit is uniform circular motion.In uniform circular motion, a body moves in a circle with constant speed but the velocity changes due to a constant change in direction.

The velocity - time graph of a body with a constant acceleration is:

A.

Straight line parallel to velocity axis

B.

Parabola

C.

Hyperbola

D.

Straight line inclined at some angle with respect to time axis

#### SOLUTION

Solution : D Since acceleration is constant, the slope of the velocity-time graph should be a straight line with constant slope. Thus, the graph which shows a straight line inclined at some angle with respect to time axis represents the velocity-time graph of a body with a constant acceleration.

A car covers a distance of 300 m in 20 seconds. Calculate the speed of the car.

A. 25 ms1
B. 20 ms1
C. 10 ms1
D. 15 ms1

#### SOLUTION

Solution : D

Given, distance travelled by the car =300 m and the time taken =20 s.
We know, speed=distancetime
speed=30020 = 15 ms1.
Speed of car is  15 ms1