Motion Test

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Motion Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?

A
x < 0, v < 0, a > 0

B
x > 0, v < 0, a < 0

C
x > 0, v < 0, a > 0

D
x > 0, v > 0, a < 0

Solution

Displacement is downward, x<0
Velocity is also downward ,v<0
But acceleration is in direction of velocity ,a>0
Question 2
1 / -0

The distances travelled by a body starting from rest and travelling with uniform acceleration in successive intervals of time each of one second will be in the ratio:

A
$$1:2:3$$

B
$$1:2:4$$

C
$$1:3:5$$

D
$$1:5:9$$

Solution

Distance traveled (s) is given by :
$$s = ut + \dfrac {1}{2} at^2$$

Now, interval wise distance traveled are:

$$s_1 - s_0 = \dfrac {a}{2} (1)^2 = \dfrac {a}{2}$$

$$s_2 - s_1 = \dfrac {a}{2} (2)^2 - \dfrac {a}{2} (1)^2 = \dfrac {3a}{2}$$

$$s_3 - s_2 = \dfrac {a}{2} (3)^2 - \dfrac {a}{2} (2)^2 = \dfrac {5a}{2}$$

Hence, its in the ratio $$ = \dfrac {a}{2} : \dfrac {3a}{2} : \dfrac {5a}{2}$$$$= 1 : 3 : 5 $$
Question 3
1 / -0

The figure shows four graphs of displacement ($$x$$) versus time ($$t$$), the graph that shows a constant, positive, and non-zero velocity is:

A
$$(a)$$

B
$$(b)$$

C
$$(c)$$

D
$$(d)$$

Solution

When an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance traveled by the object is directly proportional to the time taken. Thus, for uniform speed, a graph of distance traveled against time is a straight line.

$$Graph\ (a)-$$ The graph is a straight line and displacement is increasing with increasing time. Therefore, velocity is uniform (constant), increasing (positive), and non-zero.

$$Graph\ (b)-$$ The graph is a straight line but displacement is constant with increasing time. Therefore, velocity is zero.

$$Graph\ (a)-$$ The graph is a not straight line. Therefore, velocity is non-uniform and non-zero.
$$Graph\ (d)-$$ The graph is a straight line and displacement is decreasing with increasing time. Therefore, velocity is uniform (constant), decreasing (negaitive), and non-zero.
Question 4
1 / -0

If north is the positive direction, which situation involves positive velocity and negative acceleration?

A
A car speeding up as it moves southward

B
A car moving southward at a constant speed

C
A car slowing down as it moves northward

D
A car slowing down as it moves southward

Solution

The rate of change of velocity with respect to time is known as acceleration. Negative acceleration means a decrease in velocity with respect to time.
If a car is moving northward and slowing down, the displacement will be along the north (positive), so the velocity will be positive until it gets stopped. As the car is slowing down, the acceleration will be negative.
Question 5
1 / -0

A body reaches the ground in $$3\;s$$ when it is released with zero velocity from the top of a building. The height of the building is:

A
$$14.7\ m$$

B
$$24.4\ m$$

C
$$44.1\ m$$

D
$$66.2\ m$$

Solution

Given,
Initial velocity, $$u=0$$
Acceleration due to gravity, $$a=g=9.8\ m/s^2$$
Time taken, $$t=3\ s$$
Height of the building, $$h$$

Using Newton's second equation of motion,
$$s=ut+\dfrac{1}{2}at^2$$

$$h=0+\dfrac{1}{2}gt^2$$

$$h= \dfrac{1}{2}9.8 \times(3)^2$$

$$s= 44.1\ m$$
Question 6
1 / -0

The displacement time graph of a moving particle is shown in the figure. The instantaneous velocity of the particle is negative at the point:

A
D

B
F

C
C

D
E

Solution

The slope of displacement-time graph gives velocity.

Clearly at point E, the slope is negative. Hence, the instantaneous velocity is negative at that point.
Question 7
1 / -0

A balloon starts from rest, moves vertically upwards with an acceleration $$\dfrac{g}{8}\ \text{ms}^{-2}$$. A stone falls from the balloon after $$8\ \text{s}$$ from the start. The time taken by the stone to reach the ground is $$\left(g=9.8\ \text{ms}^{-2}\right)$$:

A
$$4\; s$$

B
$$8\; s$$

C
$$2$$$$\sqrt{2}$$ s

D
$$12\; s$$

Solution

For balloon, distance traveled by it in 8 sec,

$$s = ut + \dfrac {1}{2} gt^{2} $$ where $$u=0$$

$$s$$$$=$$$$\dfrac{1}{2}\times \dfrac{g}{8} t^2$$$$=4g$$

So, time taken by stone,

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

$$4g = \dfrac {1}{2} gt^2$$ at this height the velocity of stone will be zero

$$t^2 = \sqrt 8 \Rightarrow t = 2 \sqrt 2\ s.$$
Question 8
1 / -0

A pebble is thrown vertically upwards from a bridge with an initial velocity of $$4.9 \;ms^{-1}$$. It strikes the water after $$2\;s$$. Height of the bridge is:

A
$$19.6$$ m

B
$$14.7$$ m

C
$$9.8$$ m

D
$$4.9$$ m

Solution

Option C is correct
$$s = ut + \dfrac {1}{2} at^2$$
$$u = + 4.9 m/s$$
$$g = -9.8 m/s^2$$
$$t = 2 sec.$$
$$\Rightarrow s = 4.9 \times 2 - \dfrac {1}{2} \times 9.8 \times (2)^2 = -9.8 m$$
Height is $$|s| = 9.8\ m$$
Question 9
1 / -0

A body thrown vertically upwards reaches the highest point in $$2\;s$$. Initial velocity of projection is_____

A
9.8 $$ms^{-1}$$

B
19.6 $$ms^{-1}$$

C
29.4 $$ms^{-1}$$

D
39.2 $$ms^{-1}$$

Solution

$$v = u + at; a = -g$$
$$v = u - gt$$
$$v = 0, u = gt$$
$$u = 9.8 \times 2 = 19.6\ m/s$$
Question 10
1 / -0

A body thrown up with some initial velocity reaches a maximum height of $$50\;m$$. Another body with double the mass thrown up with double the initial velocity will reach a maximum height of :

A
$$100$$ m

B
$$200$$ m

C
$$400$$ m

D
$$50$$ m

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Re-Attempt Weekly Quiz Competition

Motion Test - 30

Result

Motion Test - 30

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SHARING IS CARING

Question 1

x < 0, v < 0, a > 0

x > 0, v < 0, a < 0

x > 0, v < 0, a > 0

x > 0, v > 0, a < 0

Solution

Question 2

$$1:2:3$$

$$1:2:4$$

$$1:3:5$$

$$1:5:9$$

Solution

Question 3

$$(a)$$

$$(b)$$

$$(c)$$

$$(d)$$

Solution

Question 4

A car speeding up as it moves southward

A car moving southward at a constant speed

A car slowing down as it moves northward

A car slowing down as it moves southward

Solution

Question 5

$$14.7\ m$$

$$24.4\ m$$

$$44.1\ m$$

$$66.2\ m$$

Solution

Question 6

D

F

C

E

Solution

Question 7

$$4\; s$$

$$8\; s$$

$$2$$$$\sqrt{2}$$ s

$$12\; s$$

Solution

Question 8

$$19.6$$ m

$$14.7$$ m

$$9.8$$ m

$$4.9$$ m

Solution

Question 9

9.8 $$ms^{-1}$$

19.6 $$ms^{-1}$$

29.4 $$ms^{-1}$$

39.2 $$ms^{-1}$$

Solution

Question 10

$$100$$ m

$$200$$ m

$$400$$ m

$$50$$ m

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