Motion in A Straight Line Test

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Motion in A Straight Line Test - 32

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Question 1
1 / -0

The rate of change of velocity of an object with respect to time is called ..........

A
Momentum

B
Displacement

C
Acceleration

D
Impulse

Solution

Acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

$$\text{acceleration} = \dfrac{\text{Change in the velociity}}{\text{time}}$$
Question 2
1 / -0

How much time elapses during this interval?

A
5 s

B
20 s

C
10 s

D
1 s

Solution

Answer is C.

We select positive direction for our coordinate system to be the direction of the velocity and choose the origin so that $$x_i\, =\, 0$$ when the braking begins; Then the,initial velocity is $$u_x\, =\, + 25$$ m/s at t = 0, and the final velocity and position are $$v_x\, =\, + 15$$ m/s and x = 200 m at time t.
Since the acceleration is constant, the average velocity in the interval can be found from the average of the initial and final velocities.
Therefore, $$v_{av.x}\, =\, \displaystyle \frac{1}{2}(u_x\, +\, v_x)\, =\, \displaystyle \frac{1}{2}(15\, +\, 25)\, =\, 20m/s$$.
The average velocity can also be expressed as
$$V_{av.x}\, =\, \displaystyle \frac{\Delta x}{\Delta t}$$. With $$\Delta x$$ = 200 m
and $$\Delta t\, =\, t\, -\, 0$$, we can solve for t:
$$t\, =\, \displaystyle \frac{\Delta x}{v_av.x}\, =\, \displaystyle \frac{200}{20} \, =\, 10s$$.
Hence, the time elapses during this interval is 10 s.
Question 3
1 / -0

A car of mas sm starts moving so that its velocity varies according to the law $$v = \beta\, \sqrt{s}$$ where $$\beta$$ is a constant, ands is the distance covered. The total work performed by all the forces which are acting on the car during the first t seconds after the beginning of motion is

A
$$m\beta^{4}t^{2}/8$$

B
$$m\beta^{2}t^{4}/8$$

C
$$m\beta^{4}t^{2}/4$$

D
$$m\beta^{2}t^{4}/4$$

Solution

$$v^2=u^2+2as$$
$$\beta^2s=0+2as$$
$$a=\dfrac{\beta^2}{2}$$
$$s=ut+0.5at^2$$
$$s=0+0.5(\beta^2/2)t^2$$
Work=mas
W=$$m(\beta^2/2)(0.5(\beta^2/2)t^2)$$
W=$$\dfrac{m\beta^4 t^2}{8}$$
Question 4
1 / -0

An observer finds the magnitudes of the acceleration of two bodies to be the same. This necessary implies that the two bodies.

A
are at rest with respect to each other

B
are at rest or move with constant velocities with respect to each other

C
are accelerated with respect to each other

D
may be at rest, moving with constant velocities or accelerated with respect to each other

Solution

Case 1 :
Both the bodies are moving with same velocity. Thus they are rest w.r.t each other but they have same acceleration.

Case 2 :
Both the bodies are moving with different constant velocities. Hence both are moving with constant w.r.t to each other and yet they can have the same acceleration.

Case 3 :
Both are moving with different velocities but in opposite direction with same acceleration.
Acceleration of 1 w.r.t 2, $$a_{12} = a - (-a) =2a$$
Hence both are accelerating w.r.t each other, although having same acceleration.
Option D is correct.
Question 5
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A block of mass m is suspended by a light thread from an elevator.The elevator is accelerating upward with uniform acceleration a. The work done by tension on the block during t seconds is (1 = 0) :

A
$$\displaystyle \frac{m}{2} (g + a) at^{2}$$

B
$$\displaystyle \frac{m}{2} (g - a) at^{2}$$

C
$$\displaystyle \frac{m}{2} gat^{2}$$

D
0

Solution

T acting on string due to mass= +mg
T acting on string due to upward accelration= +ma
Total T= m(g+a)
$$S=ut+ 0.5at^2= 0.5at^2$$
$$W= (0.5at^2)(m(g+a))$$
$$W=\dfrac{m(g+a)at^2}{2}$$
Question 6
1 / -0

A truck increases its speed from $$10km/h$$ to $$50 km/h$$ in $$20$$ seconds. Its acceleration is:

A
$$0.55 m/s^2$$

B
$$2.55 m/s^2$$

C
$$0.75 m/s^2$$

D
$$8.65 m/s^2$$

Solution

Given: Initial speed, $$u=10\ km/h=10\times \dfrac{5}{18}m/s$$

Final speed, $$v=50\ km/h=50\times \dfrac{5}{18}m/s$$

Using equation of motion,

Acceleration, $$a=\dfrac{v-u}{t}$$

$$=\dfrac{\dfrac{5}{18}\times(50-10)}{20}$$

$$=\dfrac{5}{18}\times 2$$

$$=0.55\ m/s^2$$
Question 7
1 / -0

A car is moving with a uniform velocity of 40 km h$$^{-1}$$ in straight line. Its acceleration after 1 hour is:

A
40 km h$$^{-1}$$

B
20 km h$$^{-1}$$

C
30 km h$$^{-1}$$

D
zero

Solution

Answer is D.

When the car is moving with uniform velocity, then there is nil acceleration in the motion of the car.
Therefore, the total acceleration of the car is 0.
Question 8
1 / -0

The speed of the particle moving along with straight line become half after every next second. The initial speed is $$V_0$$. The total distance travelled by the particle will be

A
$$V_0$$

B
$$2V_0$$

C
$$\infty $$

D
None of the above

Solution

Distance = $$V_0 \times 1+\dfrac{V_0}{2}\times 1+\dfrac {V_0}{4}\times 1+.....$$

$$ = V_0 \times \left (1+\dfrac{1}{2}+\dfrac{1}{4}+...... \right ) = 2V_0$$
Question 9
1 / -0

A rocket is launched to travel vertically upward with a constant velocity of 20 m/s. After travelling 35 s the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body, the height achieved it is

A
600 m

B
720 m

C
800 m

D
820 m

Solution

For upward motion of rocket with uniform velocity $$u=20ms^{-1} ; t=35s ; $$
$$h=ut=20\times35=700m$$
For upward motion after fuel cut off $$u=20ms^{-1} ;v=0 ; g=-10ms^{-2}$$
using formula $$v=u+gt$$
$$0=20-10\times t$$
$$t=2s$$
from formula $$v^2=u^2+2gh$$
putting values,we get,
$$0=20^2-2\times10\times h$$
$$0=400-20h$$
$$h=\frac{400}{20}=20m$$
Total height achieved $$=700+20=720m$$
hence,option b is correct.
Question 10
1 / -0

The dispIacement of a body is given by $$2s\, =\, gt^2$$ where g is a constant. The velocity of the body at any time t is:

A
$$gt$$

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

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

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

Solution

Given that:

$$2s = gt^{2}$$

$$s = \dfrac{1}{2} gt^{2}$$

Comparing this with the equation:

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

we get,

$$u = 0\ \& \ a=g$$

$$v = u + at = 0 + gt$$

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Motion in A Straight Line Test - 32

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Motion in A Straight Line Test - 32

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SHARING IS CARING

Question 1

Momentum

Displacement

Acceleration

Impulse

Solution

Question 2

5 s

20 s

10 s

1 s

Solution

Question 3

$$m\beta^{4}t^{2}/8$$

$$m\beta^{2}t^{4}/8$$

$$m\beta^{4}t^{2}/4$$

$$m\beta^{2}t^{4}/4$$

Solution

Question 4

are at rest with respect to each other

are at rest or move with constant velocities with respect to each other

are accelerated with respect to each other

may be at rest, moving with constant velocities or accelerated with respect to each other

Solution

Question 5

$$\displaystyle \frac{m}{2} (g + a) at^{2}$$

$$\displaystyle \frac{m}{2} (g - a) at^{2}$$

$$\displaystyle \frac{m}{2} gat^{2}$$

0

Solution

Question 6

$$0.55 m/s^2$$

$$2.55 m/s^2$$

$$0.75 m/s^2$$

$$8.65 m/s^2$$

Solution

Question 7

40 km h$$^{-1}$$

20 km h$$^{-1}$$

30 km h$$^{-1}$$

zero

Solution

Question 8

$$V_0$$

$$2V_0$$

$$\infty $$

None of the above

Solution

Question 9

600 m

720 m

800 m

820 m

Solution

Question 10

$$gt$$

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

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

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

Solution

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