Motion Test

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Motion Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

A car travels from rest with a constant acceleration $$'a'$$ for $$t$$ seconds. What is the average speed of the car for its journey, if the car moves along a straight road ?

A
$$v=\dfrac{at^2}{2}$$

B
$$v=2at^2$$

C
$$v=\dfrac{at}{2}$$

D
None

Solution

Initial speed of car $$u = 0$$
Distance covered by car in $$t$$ seconds is given by second equation of motion,
$$S = ut+\dfrac{1}{2}at^2$$
$$\implies \ S = \dfrac{1}{2}at^2$$ $$(\because u=0)$$

Average speed of car = $$\dfrac{ \text{total distance}}{\text{total time} }$$
$$v_{avg} = \dfrac{S}{t} = \dfrac{\dfrac{1}{2}at^2}{t}=\dfrac{at}{2}$$

So option C is correct.
Question 2
1 / -0

Sheela takes 15 minutes from her house to reach school on her bicycle with a speed of 2 m/s. The distance between her house and the school is

A
1.8 km

B
3 km

C
1 km

D
2.5 km

Solution
Question 3
1 / -0

A scooterist covers a distance of 3 km in 5 minutes. He travels with a speed of

A
$$100 m s^{-1}$$

B
$$360 km h^{-1}$$

C
$$100 cm s^{-1}$$

D
$$600 m min^{-1}$$

Solution
Question 4
1 / -0

$$150\ m$$ long train takes $$10\ s$$ to pass a man who is going in the same direction at the speed of $$2\ kmph$$. What is the speed of the train?

A
$$52\ kmph$$

B
$$56\ kmph$$

C
$$84\ kmph$$

D
Data inadequate

Solution

Let the speed of the train be $$x\ kmph$$
Length of the train $$ = 150 m = \dfrac{150}{1000} = \dfrac{3}{20} km$$
and time taken to cross the man = $$10$$ seconds = $$\dfrac{10}{60 \times 60}$$
Relative speed = $$(x - 2)\ kmph$$
(Both are moving in same direction)
We know that,
$$\dfrac{\text{Length of the train}}{\text{Relative speed}} = $$ Time taken to cross the man
Relative speed = $$\dfrac{3 \times 360}{20 \times 1}$$
$$x - 2 = \dfrac{3 \times 360}{20 \times 1}$$
$$ x = 54 + 2 = 56\ kmph$$
Hence, speed of train is $$56\ kmph$$.
Question 5
1 / -0

Two cars $$A$$ and $$B$$ have masses $$m_{A}$$ and $$m_{B}$$ respectively, with $$m_{A} > m_{B}$$. Both the cars are moving in the same direction with equal kinetic energy. If equal braking force is applied on both, then before coming to rest.

A
$$A$$ will cover a greater distance

B
$$B$$ will cover a greater distance

C
Both will cover the same distance

D
Distance covered by them will depend on their respective velocities

Solution

Kinetic energy $$K = \dfrac{1}{2}mv^2$$
Since, $$m_A>m_B$$ and both the cars move with same kinetic energy,
$$\implies \ v_A<v_B$$
Also, both the car with retard with same acceleration because of equal braking force before coming to rest.
So, car B will cover a greater distance than car A.
Question 6
1 / -0

Which of the following is not a unit of speed?

A
Meter/second

B
Second/meter

C
Kilometer/hour

D
Meter/minute.
Question 7
1 / -0

In uniform circular motion, direction of velocity is along the _______ drawn to the position of particle on the circumference of the circle.

A
normal

B
tangent

C
can be both

D
none of these

Solution

In uniform circular motion direction of velocity is along the $$tangent$$ drawn to the position of particle on the circumference of the circle.

If we move a stone tied to a string to describe a circular path with constant speed by holding the thread at the other end, we see that on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant. This proves the velocity is tangential to the position of the particle on the circumference.
Question 8
1 / -0

A ball is travelling with uniform translatory motion. This means that

A
It is at rest

B
The path can be straight line or circular and the ball travels with uniform speed.

C
All parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.

D
The centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

Solution

In uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant.
Question 9
1 / -0

The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are as shown in the figure. Choose the incorrect statement regarding these graphs.

A
A lives closer to the school than B

B
A starts from the school earlier than B

C
A walks faster than B

D
A and B reach home at the same time.

Solution

$$\textbf{Explanation:}$$
$$\textbf{Step 1: Positions of their respective houses from school}$$

Since the slope of OP is lesser OQ, it infers that point P is closer to O than Q.

⸫ A lives closer to a school than B. Hence, (a) is true.

$$\textbf{Step 2: Their starting times from school}$$

From the graph, for line OP, x=0; t=0

For line OQ, when x=0; t has a positive finite value which means that A started from school earlier than B.

Hence, (b) is also true.

$$\textbf{Step 3: Their respective reaching times}$$

Point P and Q lie at the same vertical line on the abscissa (t) which denotes that both A and B reach their respective homes at the same time ‘t’.

Hence, (d) is also true

$$\textbf{Step 4: Their respective walking speeds}$$

The pace of the children is indicated by the slope of the x-t graph as their motion is uniform motion.
From the graph, it can be seen that the slope of OQ is greater than OP which means that pace of B is greater than A.
Hence, (c) is false.

Final Answer : $$\textbf{Hence, the correct option is (c) A walks faster B}$$
Question 10
1 / -0

The figure shows the $$x-t$$ plot of a particle in one-dimensional motion. Two different equal intervals of time are shown. Let $$v_1$$ and $$v_2$$ be speed in time intervals $$1$$ and $$2$$ respectively. Then

A
$$v_1 > v_2$$

B
$$v_2 > v_1$$

C
$$v_1 = v_2$$

D
Data is insufficient

Solution

We know,
The slope of the $$x-t$$ graph represents the speed. The average speed of a particle shown in the $$x-t$$ graph is obtained from its slope in a particular interval of time. So, the higher the slope of the graph, the higher is the average speed.
Now see the graph,
It is clear from the graph that the magnitude of the slope is more in interval-$$1$$ than in the interval-$$2$$.
Therefore, the speed of the particle in $$2$$ will be less than the $$1$$.
Thus $$v_{1} > v_{2}$$

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Motion Test - 63

Result

Motion Test - 63

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SHARING IS CARING

Question 1

$$v=\dfrac{at^2}{2}$$

$$v=2at^2$$

$$v=\dfrac{at}{2}$$

None

Solution

Question 2

1.8 km

3 km

1 km

2.5 km

Solution

Question 3

$$100 m s^{-1}$$

$$360 km h^{-1}$$

$$100 cm s^{-1}$$

$$600 m min^{-1}$$

Solution

Question 4

$$52\ kmph$$

$$56\ kmph$$

$$84\ kmph$$

Data inadequate

Solution

Question 5

$$A$$ will cover a greater distance

$$B$$ will cover a greater distance

Both will cover the same distance

Distance covered by them will depend on their respective velocities

Solution

Question 6

Meter/second

Second/meter

Kilometer/hour

Meter/minute.

Question 7

normal

tangent

can be both

none of these

Solution

Question 8

It is at rest

The path can be straight line or circular and the ball travels with uniform speed.

All parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.

The centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

Solution

Question 9

A lives closer to the school than B

A starts from the school earlier than B

A walks faster than B

A and B reach home at the same time.

Solution

Question 10

$$v_1 > v_2$$

$$v_2 > v_1$$

$$v_1 = v_2$$

Data is insufficient

Solution

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