Motion Test

Result

Motion Test - 58

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Question 1
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Directions For Questions

The following options have dot representations. Each dot represents a time interval of one second and the motion of the ball is not necessarily horizontal.

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In which of the following option could represent the ball moving at a constant velocity?

A

B

C

D

E

Solution

For constant velocity, displacement cover by the moving particle is constant in every time interval.
Except Option A, In other options, the gap between dots is not same so that can't represent object moving with constant velocity. Whereas in Option A, Gap between dots are equal so it can represent the motion of a particle moving with constant velocity.
Question 2
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A block starts accelerating at $$5 m/s^{2}$$ from rest on a frictionless surface. Calculate the distance traveled by the block in first 4 seconds?

A
20 m

B
40 m

C
80 m

D
120 m

E
160 m

Solution

Given : $$u = 0$$ $$a = 5 m/s^2$$ $$t = 4$$ s
Using $$S = ut + \dfrac{1}{2}at^2$$
$$\therefore$$ Distance covered in 4 s $$S = 0+\dfrac{1}{2} \times 5 \times 4^2 = 40$$ m
Question 3
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The velocity-time graph below represents the velocity of a toy train as it moves north and south with velocity near the middle of the vertical axis. During which, Interval(s) is the toy train speeding up?

A
$$0\ to\ A$$ only

B
$$0\ to\ A$$ and $$D\ to\ E$$

C
$$A\ to\ B$$

D
$$B\ to\ D$$ only

E
$$A\ to\ B$$ and $$D\ to\ E$$

Solution

The toy train will speed up if the rate of change of velocity will increases with respect to time.
Therefore, the train will speed up in the intervals $$0\ to A$$ and $$D\ to\ E$$.
Question 4
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A ball of mass m is thrown straight upward from the top of a multi-storey building with an initial velocity of +15 m/s.
Find out the approximate maximum height reached by the ball?

A
23m

B
15m

C
11m

D
8m

E
5m

Solution

Given : $$u = 15$$ m/s $$a = -g = -10 m/s^2$$ (taking upward direction to be positive)
Final velocity of the ball at the highest point $$v = 0$$ m/s
Using $$v^2 - u^2 = 2aS$$
$$\therefore$$ $$0 - (15)^2 = 2 \times (-10) S$$ $$\implies S = 11.25 \approx 11$$ m
Question 5
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A block of mass 2 kg falls from the wall at a height of 2.3 m. Calculate the velocity just before it hits the ground. ($$g=-9.8 \frac{m}{s^2}$$)

A
$$-6.76 \frac{m}{s}$$

B
$$-26.86 \frac{m}{s^2}$$

C
$$45.08 \frac{m}{s}$$

D
$$13.43 \frac{m}{s}$$

E
$$3.36 \frac{m}{s}$$

Solution

Given :
$$u = 0\ ms^{-1}$$; initial velocity
$$a = -9.8\ ms^{-2}$$; acceleration due to gravity
$$h = 2.3\ m$$; height of the wall
Speed with which its hits the ground.
Apply third equation of motion-
$$v^2 = u^2 + 2gh$$
$$\Rightarrow v^2 = 0 + 2gh$$
$$\Rightarrow v = \sqrt{2gh}$$
$$\Rightarrow v = \sqrt{2 \times 9.8 \times 2.3} $$
$$ \Rightarrow v =6.76\ ms^{-1} $$ m/s
As the downward direction is taken to be a negative direction, thus the hitting velocity is $$-6.76\ ms^{-1}$$ m/s.
Question 6
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A car starts from rest and accelerates on a straight horizontal track at $$5 {m}/{{s}^{2}}$$. Find out the distance travel by the car in $$8$$ seconds?

A
$$40 m$$

B
$$60 m$$

C
$$80 m$$

D
$$120 m$$

E
$$160 m$$

Solution

Given : $$u = 0$$ m/s $$t = 8$$ s $$a = 5$$ $$m/s^2$$
Using : $$S =ut + \dfrac{1}{2}at^2$$
$$\therefore$$ $$S =0 + \dfrac{1}{2} \times 5 \times 8^2 =160\ m$$
Question 7
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The graphs above show that position versus time for three different cars: $$1, 2,$$ and $$3$$. Rank these cars according to the magnitudes of their velocities, greatest first.

A
$$1, 2, 3$$

B
$$3, 2, 1$$

C
$$2, 1, 3$$

D
$$2, 3, 1$$

E
All tie

Solution

The slope of the position-time graph gives us the speed of the object at that point. In the given graph, the slope for cars $$1,2,3$$ is $$0$$. Hence, the speed of all cars is same and is equal to $$0$$.

Hence, the correct option is D.
Question 8
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A frog that jumps upwards is under the influence of gravity and accelerates at a constant rate.
If it has an initial upwards velocity of $${v}_{0}=5.00{m}/{s}$$, approximately how far above the ground will it be when it has velocity $$v=-2.50{m}/{s}$$?

A
$$0.13m$$

B
$$0.96m$$

C
$$1.28m$$

D
$$1.59m$$

E
$$1.91m$$

Solution

Initial velocity of the frog $$v_o = 5.00$$ $$m/s$$ (Taking upward direction to be positive)
Final velocity of the frog $$v = -2.50$$ $$m/s$$
Acceleration of the frog $$a = -g = -9.8 m/s^2$$
Using 3rd equation of motion, $$v^2 - v_o^2 =2aS$$
$$\therefore$$ $$(-2.50)^2 - (5.00)^2 = 2 \times (-9.8) \times S$$ $$\implies$$ $$S = 0.96$$ $$m$$
Question 9
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The position-time graphs above represent the motions of cars $$1$$ to $$5$$.
How do they rank, according to their speeds (greatest first)?

A
$$1, 2, 3, 4, 5$$

B
$$1, 2, 3$$ and $$5$$ tie $$4$$

C
$$1, 2, 4, 3$$ and $$5$$ tie

D
$$3$$ and $$5$$ tie, $$4, 2, 1$$

E
$$5, 4, 3, 2, 1$$

Solution

The slope of position-time graph gives us the speed of the object at that point.

From the given graph, we see that the slope of cars $$3$$ and $$5$$ is zero. Hence, their speed is zero. Also, the slope of cars in decreasing order is given as $$1,2,4$$.

Hence, the correct rank according to their speeds is $$1,2,4, 3$$ and $$5$$ tie
Question 10
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A cart begins from rest at the top of a long incline and rolls with a constant acceleration of $$2m/s^{2}$$. How far has the cart moved along the incline after rolling for $$3$$ seconds?

A
$$3$$ meters

B
$$6$$ meters

C
$$9$$ meters

D
$$18$$ meters

Solution

Given:
Initial velocity of cart $$u =0 m/s$$
Acceleration of the cart $$a = 2$$ $$m/s^2$$
Distance covered by cart in 3 seconds is given second equation of motion,
$$S = ut + \dfrac{1}{2}at^2$$ where $$t =3$$ seconds
$$\therefore$$ $$S = 0 + \dfrac{1}{2 } \times 2 \times 3^2 = 9$$ meters