# Free Motion 01 Practice Test - 9th Grade

### Question 1

A coin is thrown in a vertically upward direction with a velocity of 5 ms−1. If the acceleration of the coin during its motion is 10 ms−2 in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?

#### SOLUTION

Solution :D

Given, initial velocity, u=5 ms−1, acceleration, a=−10 ms−2.

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=5−10t

⇒t=0.5 s

∴ Time taken by the coin is 0.5 s.

### Question 2

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?

#### 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.

### Question 3

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 s−2 . Find how far the train will go before it is brought to rest.

600 m

725 m

625 m

500 m

#### SOLUTION

Solution :C

Given,

initial velocity, u=90 kmph = 90×518 =25 ms−1

acceleration, a= –0.5 ms−2

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.

### Question 4

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

True

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.

### Question 5

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.

#### 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 ms−1

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 ms−1 and the time it will take to cross the bridge is 50 s.

### Question 6

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.

#### 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

### Question 7

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?

#### 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 hr−1

=1099 kmph

### Question 8

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

#### 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.

### Question 9

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

Straight line parallel to velocity axis

Parabola

Hyperbola

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.

### Question 10

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

#### SOLUTION

Solution :D

Given, distance travelled by the car =300 m and the time taken =20 s.

We know, speed=distancetime

⇒ speed=30020 = 15 ms−1.

∴ Speed of car is 15 ms−1