Free Motion 01 Practice Test - 9th Grade

Question 1

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

s = 1.5 m, t = 0.25 s

s = 0.25 m, t = 1.5 s

s = 1.25 m, t = 0.75 s

s = 1.25 m, t = 0.5 s

SOLUTION

Solution : D

Given, initial velocity, $u = 5 m s^{- 1}$ , acceleration, $a = - 10 m s^{- 2}$ .

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, $v^{2} = u^{2} + 2 a s,$
$0 = 5^{2} + [2 \times (- 10) \times s]$
$\Rightarrow s = 1.25 m$
Hence the height attained by the coin is $1.25 m$ .

Let the time taken be $t$ .
Using the first equation of motion,
$v = u + a t$
$0 = 5 - 10 t$
$\Rightarrow t = 0.5 s$
$∴$ 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?

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 $= \sqrt{(B C^{2} + A C^{2})}$
$= \sqrt{(3^{2} + 4^{2})} m = 5 m$ .

Question 3

An express train going towards Delhi is travelling at a speed of $90 k m p h$ . Brakes are applied so as to produce a uniform acceleration of $- 0.5 m s^{- 2}$ . 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, $u = 90 k m p h$ = $90 \times \frac{5}{18}$ $= 25 m s^{- 1}$
acceleration, $a = - 0.5 m s^{- 2}$
final velocity, $v = 0$ (brakes are applied)
Let 's' be the distance travelled.
From the third equation of motion,
$v^{2} = u^{2} + 2 a s$

$0 = 25^{2} + (2 \times - 0.5 \times s)$

$\Rightarrow s = 625 m$

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.
$Slope = \frac{Change in y-coordinate}{Change in x-coordinate}$
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.

13 m s^{- 1}

40 s

25 m s^{- 1}

30 s

12 m s^{- 1}

50 s

15 m s^{- 1}

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 $= 540 m$

$Speed of the train, s = \frac{D}{T}$

$Speed of the train, s = \frac{60}{5} = 12 m s^{- 1}$

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

$∴ Time taken = \frac{d}{s}$

$Time taken = \frac{600}{12} = 50 s$

$∴$ Speed of the train is $12 m s^{- 1}$ 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.

\frac{u + v}{2}

\frac{u v}{2}

2 u v

u v

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 $\frac{T}{2}$ and the next half of the journey be $\frac{T}{2}$
Let the average speed be $V_{a v}$
We know, $s p e e d = \frac{D i s t a n c e}{t i m e}$
$∴$ $D i s t a n c e = s p e e d \times t i m e$

During the first half of the journey, distance, $d_{1} = u \times \frac{T}{2}$
During the second half of the journey, distance, $d_{2} = v \times \frac{T}{2}$
Total distance $= d = d_{1} + d_{2}$
$= u \times$ $\frac{T}{2} + v \times \frac{T}{2}$
$\Rightarrow d = \frac{T}{2} (u + v)$

Average speed, $V_{a v}$ =   $\frac{T o t a l d i s t a n c e}{T o t a l t i m e t a k e n}$
   $= \frac{d}{T}$
$= \frac{\frac{T}{2} (u + v)}{T}$
$= \frac{u + v}{2}$

Question 7

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

4099 kmph

3099 kmph

1099 kmph

2099 kmph

SOLUTION

Solution : C

Given, the radius of the orbit, $R = 4200 k m$
The time taken to revolve around the Earth, $T = 24 h r (1 d a y)$
The speed of the satellite, $v = \frac{Circumference of the orbit}{Time taken}$
$= \frac{2 π R}{T}$
$= \frac{2 \times 3.14 \times 4200 km}{24 hr}$
$= 1099 km {hr}^{- 1}$
$= 1099 kmph$

Question 8

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.

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.

25 m s^{- 1}

20 m s^{- 1}

10 m s^{- 1}

15 m s^{- 1}

SOLUTION

Solution : D

Given, distance travelled by the car $= 300 m$ and the time taken $= 20 s$ .
We know, $s p e e d = \frac{d i s t a n c e}{t i m e}$
$\Rightarrow s p e e d = \frac{300}{20}$ = $15 m s^{- 1}$ .
$∴$ Speed of car is $15 m s^{- 1}$

Free Motion 01 Practice Test - 9th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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Force and Laws of Motion 01
Work and Energy 01
Gravitation 01