Motion Test

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Motion Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

A particle moves on the $$x-$$ axis. When the particle's acceleration is positive and increasing then

A
It must speeding up

B
It must be slowing down

C
It may speeding up

D
None of the these
Question 2
1 / -0

A scooterist travels at 30 km/h along straight path for 20 min. What is distance?

A
1.5 km

B
6 km

C
10 km

D
90 km

Solution

If the scooter is travelling at precisely $$30 km/h$$ in a perfectly straight line.
We know $$1$$ hour=$$60$$ minutes
Therefore to calculate the distance traveled in the started fraction of the hour.
$$\therefore 20$$ minutes is $$ = \frac{{20}}{{60}} = \frac{1}{3}$$ of an hour or $$3.3333$$%
$$30\,km \times \frac{1}{3} = 10\,km$$
Therefore $$10km$$ is the distance traveled in $$20$$ minutes of travelling at a speed of $$30 km/h$$
Question 3
1 / -0

Particle is dropped from the height of 20m on horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes 6m$${ s }^{ -2 }$$. Find the horizontal displacement of the particle till it reaches ground.

A
6m

B
10m

C
12m

D
24m

Solution

The correct opption is C.

Given,

$$ height =20m$$

$$Acceleration=6m/s^2$$

We know $$u=0,g=10m/s^2,s=20m$$

So,

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

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

$$20=\dfrac{1}{2}10t^2$$

$$t=2s$$

Since $$a=6m/s^2$$ in the horizonttal,

Thus,

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

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

$$=\dfrac{1}{2}g\times2^2$$

$$=12m$$
Question 4
1 / -0

A ball is dropped downwards. After 2 second another ball is dropped downwards from the same point. What is the distance between them after 3 second?

A
25 m

B
20 m

C
50 m

D
9.8 m

Solution
Question 5
1 / -0

A particle moves for $$8\ s$$. It first accelerates from rest and then retards to rest. If the retardation be $$3$$ times the acceleration, then the time for which it accelerates will be

A
$$2s $$

B
$$3s$$

C
$$4s$$

D
$$6s$$

Solution

Let $$t$$ be the time of acceleration
Then retardation time $$ = 8 - t$$
total time $$= 8\ s$$
Let $$V_1$$ be velocity after acceleration
let acceleration be $$a$$
$$\begin{array}{l} { V_{ 1 } }=u+at \\ here\, \, u=0 \\ { V_{ 1 } }=at\to \left( i \right) \end{array}$$
none retardation $$= 3a$$ (given)
Hence,
$$\begin{array}{l} V={ V_{ 1 } }-3a\left( { 8-t } \right) & \\ V=0, & as\, \, body\, \, comes\, \, to\, \, rest \\ { V_{ 1 } }=3a\left( { 8-t } \right) & \\ at=3a\left( { 8-t } \right) ,put\, \, { V_{ 1 } }\, \, from\, \, \left( i \right) & \\ t=24-3t & \\ 4t=24 & \\ t=6\, \, { { second } } & \end{array}$$
Question 6
1 / -0

A bullet initially moving with a velocity $$20ms^{-1}$$ strikes a target and comes to rest after penetrating a distance $$10 cm $$ in the target .Calculate the retardation caused by the target.

A
$$1000m/s^2$$

B
$$2000m/s^2$$

C
$$3000m/s^2$$

D
$$4000m/s^2$$

Solution

Given,
$$u=20m/s$$
$$s=10cm=0.1m$$
$$v=0m/s$$
From 3rd equation of motion,
$$2as=v^2-u^2$$
$$a=\dfrac{v^2-u^2}{2s}=\dfrac{0^2-20\times 20}{2\times 0.1}$$
$$a=-2000m/s^2$$
Negative sign means it is retardation.
The correct option is B.
Question 7
1 / -0

A train starting from a railway station and moving with uniform acceleration, attain a speed of $$40km {h^{- 1}}$$ in $$10$$ minutes. Its acceleration is:

A
$$18.5m {s^{- 2}}$$

B
$$1.85cm {s^{- 2}}$$

C
$$18.5cm {s^{- 2}}$$

D
$$1.85m {s^{- 2}}$$

Solution

$$acc = \dfrac{{{v_2} - {v_1}}}{t} = \dfrac{{{v_2} - 0}}{t} = \dfrac{{40 - 0}}{{10 \times 60}}$$

$$ = \dfrac{{40 \times 1000 \times 100}}{{3600 \times 10 \times 60}}cm{s^{ - 2}}$$

$$ = 1.85cm{s^{ - 2}}$$
Question 8
1 / -0

The average speed of an object is defined to be

A
One half of the sum of the maximum and the minimum speeds

B
Distance it travels multiplied by the time it takes

C
The distance it travels divided by the time it takes

D
The speed determined over an infinitesimally small time interval

E
The value of the speed at the midpoint of the time interval

Solution

The average speed of an object is defined as the distance traveled divided by the time
$$v=\dfrac{D}{t}$$

Where, $$v$$ = speed,
$$ D$$ = distance and
$$t$$= time
Question 9
1 / -0

What determine the nature of the path followed by the particle

A
Speed

B
Velocity

C
Acceleration

D
None

Solution

Without initial velocity and acceleration path of the particle cannot be decided. So at a time velocity and acceleration are required to determine the path followed by the particle.
Question 10
1 / -0

A body starts from the rest with uniform acceleration. If its displacement in the 3rd seconds and 7th second are $$x_1$$ and $$x_2$$, then:

A
$$13x_1 = 5 x_2$$

B
$$5x_1 = 13 x_2$$

C
$$3x_1 = 7 x_2$$

D
$$7x_1 = 3 x_2$$

Solution

$$s_n$$ is the distance covered in $$n{th}$$ sec
$$s_n=u+\dfrac{a}{2}(2n-1)$$
$$u=0$$
$$\dfrac{x_1}{x_2}=\dfrac{2\times 3-1}{2\times 7-1}$$
$$13x_1=5x_2$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 65

Result

Motion Test - 65

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Rank

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SHARING IS CARING

Question 1

It must speeding up

It must be slowing down

It may speeding up

None of the these

Question 2

1.5 km

6 km

10 km

90 km

Solution

Question 3

6m

10m

12m

24m

Solution

Question 4

25 m

20 m

50 m

9.8 m

Solution

Question 5

$$2s $$

$$3s$$

$$4s$$

$$6s$$

Solution

Question 6

$$1000m/s^2$$

$$2000m/s^2$$

$$3000m/s^2$$

$$4000m/s^2$$

Solution

Question 7

$$18.5m {s^{- 2}}$$

$$1.85cm {s^{- 2}}$$

$$18.5cm {s^{- 2}}$$

$$1.85m {s^{- 2}}$$

Solution

Question 8

One half of the sum of the maximum and the minimum speeds

Distance it travels multiplied by the time it takes

The distance it travels divided by the time it takes

The speed determined over an infinitesimally small time interval

The value of the speed at the midpoint of the time interval

Solution

Question 9

Speed

Velocity

Acceleration

None

Solution

Question 10

$$13x_1 = 5 x_2$$

$$5x_1 = 13 x_2$$

$$3x_1 = 7 x_2$$

$$7x_1 = 3 x_2$$

Solution

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