Motion in A Straight Line Test

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Motion in A Straight Line Test - 17

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Question 1
1 / -0

A constant force acting on a body of mass of 5kg change its speed from 5 $$ms^{-1}$$ to 10 $$ms^{-1}$$ in 10 s without changing the direction of motion. The force acting on the body is

A
1.5 N

B
2 N

C
2.5 N

D
5 n

Solution

Here, m=5 kg, u=5 $$ms^{-1}$$, v=10 $$ms^{-1}$$, t= 10s
Using v=u+at
$$a=\dfrac{v-u}{t}=\dfrac{(10-5)ms^{-1}}{10 s}= 0.5 ms^{-2}$$
As F=ma
$$\therefore F=(5 kg)(0.5ms^{-2})= 2.5N$$
Question 2
1 / -0

The motion of particle of mass m is given by $$y=ut+\frac{1}{2}gt^2$$. The force acting on the particle is

A
mg

B
$$\frac{mu}{t}$$

C
2mg

D
$$\frac{2mu}{t}$$

Solution

Here $$y=ut+\dfrac{1}{2}gt^2$$.
$$\therefore, v=\dfrac{dy}{dt}=u+gt$$
Acceleration, $$a=\dfrac{dv}{dt}=g$$
So, the net force acting on the particle is, F=ma=mg
Question 3
1 / -0

What is the acceleration of the race car that moves at constant velocity of 300 km/hr ?

A
73.32 $$m/sec^2$$.

B
0 $$m/sec^2$$

C
63.33 $$m/sec^2$$.

D
53.33 $$m/sec^2$$.

Solution

Since the car is moving with a constant velocity, so change in the velocity of car is zero i.e. $$\Delta v = 0$$
Acceleration of car $$a = \dfrac{\Delta v}{\Delta t} = 0 \ m/s^2$$
Question 4
1 / -0

A bullet is fired from the cart vertically at the same instant cart begins to accelerate forward. Which of the following best describes the subsequent motion of the bullet?

A
The bullet goes up and then straight back down into the cart

B
The bullet goes up and lands in front of the cart

C
The bullet goes up and lands behind the cart

D
The bullet stops in the air as the cart is accelerating and "floats" until the cart stops accelerating

E
The bullet goes up and to the right of the cart.

Solution

As the bullet is fired vertically upwards, thus the bullet does not have velocity in horizontal forward direction and hence it has zero horizontal displacement.
Also the cart has acceleration in forward direction, thus the cart has finite displacement in forward direction.
Hence the bullet goes up and lands behind the cart.
Question 5
1 / -0

The displacement of a body is given by $$s=\dfrac{1}{2}gt^{2}$$, where $$g$$ is acceleration due to gravity. The velocity of the body at any time $$t$$ is

A
$$\dfrac{gt^{3}}{6}$$

B
$$\dfrac{gt^{2}}{6}$$

C
$$gt$$

D
$$\dfrac{gt}{2}$$

Solution

Given,

Displacement of body, $$s=\dfrac{1}{2}a{{t}^{2}}$$

Kinematic equation, $$s=ut+\dfrac{1}{2}a{{t}^{2}}$$

On comparing, it is clear that acceleration is equal to$$a=g\,$$and initial velocity $$u=0.$$

Apply kinematic equation

$$ v=u+at $$

$$ v=0+gt $$

$$ v=gt $$

Hence, velocity at time $$t$$ is $$gt.$$
Question 6
1 / -0

A bullet moving with a speed of $$100ms^{-1}$$ can just penetrate into two planks of equal thickness. Then the number of such planks, if speed is double will be:

A
$$6$$

B
$$10$$

C
$$4$$

D
$$8$$

Solution

Given that,

Speed of bullet $$u=100\,m/s$$

Suppose that, the thickness of one plate is s

Now, from the equation of motion

$$ {{v}^{2}}-{{u}^{2}}=-2as $$

$$ 0-{{u}^{2}}=-2as\left( s \right) $$

$$ {{u}^{2}}\propto s $$

Now,

$$ \dfrac{v_{2}^{2}}{v_{1}^{2}}=\dfrac{{{s}_{2}}}{{{s}_{1}}} $$

$$ \dfrac{\left( {{\left( 2\times 100 \right)}^{2}} \right)}{{{\left( 100 \right)}^{2}}}=\dfrac{{{s}_{2}}}{{{s}_{1}}} $$

$$ {{s}_{2}}=4{{s}_{1}} $$

$$ {{s}_{2}}=4\times 2s $$

$$ {{s}_{2}}=8s $$

Hence, the number of such planks is $$8$$
Question 7
1 / -0

A bird flies for 4 s with a velocity of $$\left| {t - 2} \right|m/s$$ in a straight
line,Where t is time in seconds.It covers a distance of

A
2 m

B
4 m

C
6 m

D
8 m

Solution

Velocity, $$u=|t-2|$$
Time, $$t=4\ s\\$$

at $$t=0,\ u=2\ m/s\\t=2,\ u=0\ m/s\\t=4\ , u=2\ m/s$$

between time $$t=0\ to\ t=2 sec$$
$$a=\dfrac{0-2}{2}=-1\ m/s^2$$
$$s_1=ut+\dfrac12 at^2\\ s_1=2\times 2+\dfrac12\times (-1)\times 4=2$$

Between time $$t=2\ to\ t=4 sec$$
$$a=\dfrac{0-2}{2}=1\ m/s^2\\s_2=\dfrac12\times 1\times 4=2$$

Thus,
Total distance;
$$s_1+s_2=2+2=4\ m$$

Thus, during the 4 seconds, the bird travels a distance of 4 m.
Question 8
1 / -0

At any instant, the velocity and acceleration of a particle moving along a straight line v and a. The speed of the particle is increasing if

A
v > 0, a > 0

B
v < 0, a > 0

C
V > 0, a < 0

D
v > 0, a = 0

Solution

acceleration of fan must be incensing in order to gain speed. if acceleration is increasing then velocity also increases.
Question 9
1 / -0

Consider a train which can accelerate with an acceleration of $$20cm/s^2$$ and slow down with deceleration of $$100cm/s^2$$. Find the minimum time for the train to travel between the stations 2.7km apart.

A
90 s

B
180 s

C
160 s

D
240 s

Solution

Max. Speed ,   $$ v = u + at = 0.2 t ,$$

                    $$ v = 0.2 t_1 $$------------(1)

      Dec --- $$v = u –at$$

                   $$ 0 = v – 1.0 t_2 ,       v = t_2 $$----(2)

$${t_1} = \dfrac{{{t_2}}}{{0.2}} = 5{t_2}\,\,\,\,$$

       $$ S = ut + \dfrac{1}{2} at^2$$

              $$S = 2700 m$$
         $$ 2700 = 0.1 [ (5t_2)^2 + (t_2)^2 - \frac{1}{2} {t_2}^2]$$

$$ r= 80 cm$$

$$ 3 { t_2}^2 = 2700 $$

              $$ t_2 = 30 sec,         t_1 = 150 sec$$
Total time = $$t_1 + t_2 = 150 + 30 = 180 sec$$
Question 10
1 / -0

A ball with a mass of 0.5 Kg is thrown vertically upward with a speed of 15 m/s. What are its speed and direction two seconds later ?

A
10 m/s upwards

B
5 m/s upwards

C
zero

D
5 m/s downward

E
10 m/s downward

Solution

Given,

Initial velocity, $$u=15m{{s}^{-1}}$$

Acceleration, $$a=-10\,m{{s}^{-2}}$$

Apply kinematic equation of motion

$$ v=u+at $$

$$ v=15-10\times 2=-5\,m{{s}^{-1}} $$

Hence, velocity after $$2\,\sec $$ is $$5\,m{{s}^{-1}}$$ downward.

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Motion in A Straight Line Test - 17

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Motion in A Straight Line Test - 17

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Question 1

1.5 N

2 N

2.5 N

5 n

Solution

Question 2

mg

$$\frac{mu}{t}$$

2mg

$$\frac{2mu}{t}$$

Solution

Question 3

73.32 $$m/sec^2$$.

0 $$m/sec^2$$

63.33 $$m/sec^2$$.

53.33 $$m/sec^2$$.

Solution

Question 4

The bullet goes up and then straight back down into the cart

The bullet goes up and lands in front of the cart

The bullet goes up and lands behind the cart

The bullet stops in the air as the cart is accelerating and "floats" until the cart stops accelerating

The bullet goes up and to the right of the cart.

Solution

Question 5

$$\dfrac{gt^{3}}{6}$$

$$\dfrac{gt^{2}}{6}$$

$$gt$$

$$\dfrac{gt}{2}$$

Solution

Question 6

$$6$$

$$10$$

$$4$$

$$8$$

Solution

Question 7

2 m

4 m

6 m

8 m

Solution

Question 8

v > 0, a > 0

v < 0, a > 0

V > 0, a < 0

v > 0, a = 0

Solution

Question 9

90 s

180 s

160 s

240 s

Solution

Question 10

10 m/s upwards

5 m/s upwards

zero

5 m/s downward

10 m/s downward

Solution

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