Motion Test

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Motion Test - 48

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Question 1
1 / -0

A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s respectively $(g = 10\ m/s^{2})$ . Find separation between them after one second.

A
5 m

B
12 m

C
3 m

D
7 m

Solution

$s_A = ut - (1/2)gt^2$
$= 5t - \displaystyle \frac{1}{2} \times 10 \times t^{2}$
$= 5 \times 1 - 5 \times 1^{2}$
$= 5 - 5 = 0\ m$

$s_B = ut - \displaystyle \frac{1}{2} gt^{2}$
$= 10 \times 1 - \displaystyle \frac{1}{2} \times 10 \times 1^{2}$
$= 10 - 5 = 5\ m$
Therefore, $s_B - s_A =\, separation\, = 5m$
Question 2
1 / -0

Figure shows the position of a particle moving on the x-axis as a function time

A
the particle has come to rest 6 times

B
the maximum speed is at t=6 s

C
the velocity remains positive for t=0 to t=6 s

D
the average velocity for the total period show in negative

Solution

Slope of the displacement-time graph gives the instantaneous velocity of the particle.
Thus the particle has zero velocity at the instant where the slope of the graph is zero i.e at points A, B, C, D, E & F
Hence option A is correct.
Question 3
1 / -0

A bullet loses $\dfrac{1}{20}$ of its velocity in passing through a plank. What is the least number of planks required to stop the bullet?

A
9

B
10

C
11

D
12

Solution

let the width of a plank be d and no of planks be n
then final speed after one plank is $v = u - \dfrac{u}{20} = \dfrac{19u}{20}$

$v^2 = u^2 +2aS$
Velocity decreasing so acceleration is negative.
$(\dfrac{19u}{20})^2 = u^2 -2a\times d$
$a=\dfrac{39u^2}{800d}$

number of planks required to stop the bullet

$V^2 = U^2 + 2aS$
$0 = u^2 - 2\times \dfrac{39u^2}{800d} \times nd$
$n = \dfrac{400}{39} = 10.25$

$\therefore \text{ There are } 11$ full planks.
Question 4
1 / -0

If an object is moving eastward and slowing down, then the direction of its velocity vector is

A
Eastward

B
Westward

C
Neither eastward nor westward

D
Not enough information to tell

Solution

The direction of the velocity vector will always be in the direction same as that of the direction in which the object moves.
So, the direction of velocity here will be eastward as the displacement is eastward.
Question 5
1 / -0

A student determined to test the law of gravity for himself walks off a sky scraper 320 m high with a stopwatch in hand and starts his free fall (zero initial velocity). 5 second later, superman arrives at the scene and dives off the roof to save the student. What must be the maximum height of the skyscraper so that even superman cannot save him.

A
65 m

B
85 m

C
125 m

D
145 m

Solution

Assuming superman can move at infinite speed, he would be able to save the student unless he has already fallen onto the pavement in 5 seconds.
Maximum height of the skyscraper so that the student falls on to the pavement in 5 seconds is:
$h = \dfrac{1}{2} gt^{2} = \dfrac{1}{2} (10) (25) = 125m$
Hence, option C is correct.
Question 6
1 / -0

Figure shows the position time graph of a particle moving on the x-axis.

A
The particle is continuously going in positive x direction

B
Area under x - t curve shows the displacement of particle

C
The velocity increases up to a time $t_0$ , and then becomes constant.

D
The particle moves at a constant velocity up to a time $t_0$ , and then stops.

Solution

Until time $t_{0}$ the graph shows a constant slope (which is non zero), which tells us that it is moving with constant velocity.
After time $t_{0}$ the position doesn't change with time. Hence the body stops.
Option D is correct.
Question 7
1 / -0

Which of the following four statement is false?

A
A body can have zero velocity and still be accelerated.

B
A body can have a constant velocity and still have a varying speed.

C
A body can have a constant speed and still have a varying velocity.

D
The direction of the velocity of a body can change when its acceleration is constant.
Question 8
1 / -0

A car falls off a bridge and drops to the ground in $0.5 s$ . Let ${g=10m/s^2}$ . How high is the bridge from the ground?

A
$1.25 m$

B
$2.5 m$

C
$5 m$

D
$10 m$

Solution

Answer is A.

In this case, the distance traveled by the car off to the ground is equal to the total height of the bridge.
So, the total distance traveled by the falling car to the ground is calculated as follows.
$s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$ where,
$s$ - distance covered
$u$ - initial velocity which is 0 in this case
$a$ - acceleration due to gravity which is $10m/{ s }^{ 2 }$
$t$ - 0.5 seconds

Substituting these values in the given equation we get,
$s=0\times 0.5+\dfrac { 1 }{ 2 } \times 10\times 0.5^{ 2 }\quad =\quad 1.25m.$

Hence, the height of the bridge is 1.25 m.
Question 9
1 / -0

Which one of the following is travelling with the fastest speed?

A
An athlete running with a speed of 10 m/s

B
A bicycle moving with a speed of 200 m/min

C
A tortoise covering 100 metres in 15 minutes

D
A motorcycle moving with a speed one km/h

Solution

To get the correct answer,we need to convert each speed in same unit i.e. m/s
A)An athlete running with a speed of 10 m/s
B)A bicycle moving with a speed of 200 m/min it can be written as
    $=\frac{200}{60}m/s$
    $=3.33m/s$
C)A tortoise covering 100 metres in 15 minutes
   $=\frac{100}{15}m/min$
   $=\frac{100}{{15}\times{60}}m/s$
   $=0.11m/s$
D)A motorcycle moving with a speed one km/h
   $=1km/h$
   $=\frac{1000}{{1}\times{60}\times{60}}$
   $=0.27m/s$
Comparing A, B, C, and D one can easily say that answer is option A i.e.
An athlete running with a speed of 10 m/s is the fastest.
Question 10
1 / -0

A stone is dropped from a height of 20 m. How long will it take to reach the ground? ( ${g=10 m/s^2}$ ).

A
2 s

B
0.5 s

C
4 s

D
1 s

Solution

Given,
$u=0m/s$
$g=10m/s^2$
$h=20m$

3rd equation of motion,
$2gh=v^2-u^2$
$2\times 10\times 20=v^2-0^2$
$v^2=400$
$v=20m/s$

1st equation of motion,
$v=u+gt$
$20=0+10t$
$t=\dfrac{20}{10}=2sec$
The correct option is A.

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Motion Test - 48

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Motion Test - 48

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Question 1

5 m

12 m

3 m

7 m

Solution

Question 2

the particle has come to rest 6 times

the maximum speed is at t=6 s

the velocity remains positive for t=0 to t=6 s

the average velocity for the total period show in negative

Solution

Question 3

9

10

11

12

Solution

Question 4

Eastward

Westward

Neither eastward nor westward

Not enough information to tell

Solution

Question 5

65 m

85 m

125 m

145 m

Solution

Question 6

The particle is continuously going in positive x direction

Area under x - t curve shows the displacement of particle

The velocity increases up to a time t0t_0t0​, and then becomes constant.

The particle moves at a constant velocity up to a time t0t_0t0​, and then stops.

Solution

Question 7

A body can have zero velocity and still be accelerated.

A body can have a constant velocity and still have a varying speed.

A body can have a constant speed and still have a varying velocity.

The direction of the velocity of a body can change when its acceleration is constant.

Question 8

1.25m1.25 m1.25m

2.5m2.5 m2.5m

5m5 m5m

10m10 m10m

Solution

Question 9

An athlete running with a speed of 10 m/s

A bicycle moving with a speed of 200 m/min

A tortoise covering 100 metres in 15 minutes

A motorcycle moving with a speed one km/h

Solution

Question 10

2 s

0.5 s

4 s

1 s

Solution

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The velocity increases up to a time $t_0$ , and then becomes constant.

The particle moves at a constant velocity up to a time $t_0$ , and then stops.

$1.25 m$

$2.5 m$

$5 m$

$10 m$