Motion in A Straight Line Test

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Motion in A Straight Line Test - 47

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Question 1
1 / -0

A bullet of mass m looses $$\left(\dfrac{1}{n}\right)^2$$ of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be:

A
$$\dfrac{n^2}{2n-1}$$

B
$$\dfrac{2n^2}{n-1}$$

C
Infinite

D
n

Solution

Let uniform resistance of 'a' acts on bullet as it passes through plank. If 'd' is thickness one plank
$$\therefore \dfrac{v^2(n-1)^2}{n^2}=v^2- 2ad$$
$$\Rightarrow a= v^2\left[1 \dfrac{(1-n)^2}{n^2}\right] =\dfrac{(2n-1)}{2dn^2}v^2$$
If N is plank are arranged side by side, then $$v^2= 2a(Nd)$$
$$\Rightarrow N=\dfrac{v^2}{2ad}=\dfrac{v^22dn^2}{2d\times (2n-1)v^2}=\dfrac{n^2}{(2n-1)}$$
Question 2
1 / -0

Newtons law are not valid in

A
Both inertial as well as non- inertial frame of reference.

B
A frame moving with constant velocity w.r.t. an inertial frame.

C
All reference frames which are at rest w.r.t. an inertial frame.

D
The reference frame attached to the earth.

Solution

Newton's Law is only valid in inertial frame of reference. If there is a non - inertial frame of reference then it won't be valid.
Hence option $$\textbf A$$ is the correct answer.
Question 3
1 / -0

Two cars $$A$$ and $$B$$ are initially $$100\ m$$ apart with $$A$$ behind $$B$$. Car $$A$$ starts from rest with a constant acceleration $$2\ m/s^{2}$$ towards car $$B$$ and at the same instant car $$B$$ starts moving with constant velocity $$10\ m/s$$ in the same direction. Time after which car $$A$$ overtakes car $$B$$ is:

A
$$14.36\ s$$

B
$$16.18\ s$$

C
$$10\ s$$

D
$$16.54\ s$$

Solution

Distance traveled by car $$A= S_A$$
Distance traveled by car $$B= S_B$$
According to formula $$S=ut+\frac {1}{2}at^2$$
$$S_A=0t+\frac {1}{2} \times 2t^2$$
$$S_B=10t+\frac {1}{2} \times 0t^2$$
Car $$A$$ will overtake car $$B$$ when
$$S_A=S_B+100$$
Putting the value $$S_A$$ and $$S_B$$
$$+2=10t+100$$
$$+2-10t-100=0$$
After solving for $$t$$, we get
$$t=16.18\, \sec$$ possible
$$t=-6.10\, \sec$$ not possible
So option B is correct.
Question 4
1 / -0

A ball dropped from the top of the tower falls first half height of the tower in 10s. The total time spent by a ball in the air is $$\left[ {Take\;g = 10m/{s^2}} \right]$$

A
$$14.14s$$

B
$$15.25s$$

C
$$12.36s$$

D
$$17.36s$$

Solution

Let the height of tower is h. Now considering the equation as

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

$$ h=ut+\dfrac{1}{2}a{{t}^{2}}...........(1) $$

$$ here,\,\,h=\dfrac{h}{2} $$

In 10 sec, object is reaching half the height of tower.

Initially the body is rest so u = 0

It is also given that a = 10 ms^-2

Time to reach first half height of tower t = 10 sec

Put the values in (1)

$$ \dfrac{h}{2}=0+\dfrac{1}{2}\times 10\times {{10}^{2}} $$

$$ h=1000\,\,m $$

The total time spent by a ball in the air is given by the relation as

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

$$ 1000=\dfrac{1}{2}\times 10\times {{t}^{2}} $$

$$ t=10\sqrt{2} $$

$$ t=14.14\,\sec $$
Question 5
1 / -0

A body moves for a total of nine second starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of acceleration and then stops. The duration of uniform acceleration

A
$$3\ s$$

B
$$4.5\ s$$

C
$$5\ s$$

D
$$6\ s$$

Solution

Consider the acceleration of the is $$\alpha $$ and retardation is $$\beta $$
and according to question retardation is twice that acceleration then
$$\beta = 2a$$

Body moves with acceleration $$\alpha $$ for $${ t_{ 1 } }$$ sec
$$v=u+at$$
$$v=0+\alpha { t_{ 1 } }$$ $$.....(1)$$

Body retards with $$2\alpha $$ for time $${ t_{ 2 } }$$ sec and then body becomes rest .
$$v=u+at$$

Here,

$$\begin{array}{l} u=\alpha { t_{ 1 } } \\ v=0 \\ a=-2\alpha \end{array}$$

$$\begin{array}{l} 0=\alpha { t_{ 1 } }-2\alpha { t_{ 2 } } \\ { t_{ 1 } }=2{ t_{ 2 } } \end{array}$$

As
$${t_1} + {t_2} = 9\sec $$

Then,
$${t_1} = 3\sec $$
$${t_2} = 6\sec $$

Therefore, $$3\;sec$$ body moves with uniform acceleration.

Hence, Option $$A$$ is the correct answer.
Question 6
1 / -0

A car starts from rest and acceleration at $$4\ {m/s}^{2}$$ for $$5\ s$$. After that it moves with constant velocity for $$25\ s$$ and then retards at $$2\ m/s^{2}$$ until it comes to rest. The total distance travelled by the car is

A
$$650\ m$$

B
$$527\ m$$

C
$$675\ m$$

D
$$573\ m$$

Solution

For $$0-5 sec$$,
$$a= 4m/s^2$$, Let velocity at $$ 5 sec$$ is $$v$$ and distance is $$S_1$$
Using $$v=u+at$$ $$\Rightarrow v= 0+4\times 5=20 m/s$$
Using $$s= ut+\dfrac{1}{2}at^2$$ $$\Rightarrow S_1=0+\dfrac{1}{2}\times 4\times 5^2=50m$$
For next $$25sec$$
$$v= 20 m/s$$
$$\Rightarrow S_2= 20\times 25= 500 m$$
For comes to rest,
$$u= 20 m/s $$ , $$a=-2 m/s^2$$
Using $$v^2=u^2+2as$$ $$\Rightarrow 0=20^2-2\times 2\times S_3$$
$$\Rightarrow S_3= 100 m$$
Total distance $$= S_1+S_2+S_3= 50+500+100=650 m$$
Question 7
1 / -0

An object starts 5m from origin and moves with an initial velocity of 5 $$m{s^{ - 1}}$$ and has an acceleration of 2 $$m{s^{ - 2}}$$. After 10 sec, the object is how far from the origin?

A
150 m

B
145 m

C
155 m

D
55 m

Solution

Displacement in $$10seconds$$ is $$ut +at^2/2$$
$$=5\times10 +2\times 10^2/2=150meter$$
Final position is initial position+displacement $$5+150=155meter$$
Question 8
1 / -0

A car travelling with a velocity of $$80 km/h$$ slowed down to $$44 km/h$$ in $$15 s$$. The retardation is

A
$$0.67m/{s^2}$$

B
$$1m/{s^2}$$

C
$$1.25m/{s^2}$$

D
$$1.5m/{s^2}$$

Solution

$$\begin{array}{l}v = u + at\\44 \times \dfrac{5}{{18}}m/s\\ = 80 \times \dfrac{5}{{18}}m/s + a \times 15\\ - 36 \times \dfrac{5}{{18}} = a \times 15\\a = \dfrac{{ - 2}}{3}m/{s^2}\\retardation = 0.67m/{s^2}\end{array}$$
Question 9
1 / -0

A train starts from rest from a station with acceleration $$0.2\ m/{s^2}$$ on a straight track and then comes to rest after attaining maximum speed on another station due to retardation $$0.4\ m/{s^2}$$. If total time spent is half an hour, then a distance between two stations is [ Neglect the length of the train]

A
$$216 Km$$

B
$$512 Km$$

C
$$728 Km$$

D
$$1296 Km$$

Solution

Given,

Velocity $${{v}_{1}}$$ achieve in time $${{t}_{1}}$$,due to acceleration, $${{a}_{1}}=0.2\,m{{s}^{-2}}$$

$$ {{v}_{1}}={{u}_{1}}+a{{t}_{1}} $$

$$ \Rightarrow {{v}_{1}}=0+0.2{{t}_{1}} $$

Velocity reduce to zero in time $${{t}_{2}}$$, due to retardation, $${{a}_{2}}=-0.4\,m{{s}^{-2}}$$

$$ {{v}_{2}}={{v}_{1}}+a{{t}_{2}} $$

$$ \Rightarrow 0={{v}_{1}}-0.4{{t}_{2}} $$

Add time

$$ {{t}_{1}}+{{t}_{2}}=\dfrac{{{v}_{1}}}{0.2}+\dfrac{{{v}_{1}}}{0.4} $$

$$ \Rightarrow 30\times 60={{v}_{1}}\times 7.5 $$

$$ \Rightarrow {{v}_{1}}=240\,m/s $$

Apply kinematic equation

$$ {{v}^{2}}-{{u}^{2}}=2as $$

$$ s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a} $$

Total distance $${{S}_{1}}+{{S}_{2}}=\dfrac{{{v}_{1}}^{2}}{2{{a}_{1}}}+\dfrac{{{v}_{1}}^{2}}{2{{a}_{2}}}=\dfrac{{{240}^{2}}}{2\times 0.2}+\dfrac{{{240}^{2}}}{2\times 0.4}=216\,km$$
Question 10
1 / -0

A body moving along the positive $$x-axis$$ with uniform acceleration of $$ - 4m{s^{ - 2}}$$. Its velocity at $$x=0$$ is $$ 10m{s^{- 1}}$$. The time taken by the body to reach a point at $$ x = 12cm$$ is:

A
$$(2s)$$

B
$$(0.25s)$$

C
$$(3s)$$

D
$$(0.0125s)$$

Solution

Acceleration $$a=-4 m/s^2$$
Initial velocity $$u=10 m/s$$
Distance traveled $$x=12 cm=12\times 10^{-2} m$$
From kinematics, $$v^2-u^2=2ax$$
$$v^2=u^2+2ax$$
$$=100+2\times (-4) \times 12 \times 10^{-2}$$
$$=100-0.96=99.04$$
$$v=9.95 m/s$$
$$v=u+at$$
$$t=\dfrac{v-u}{a}$$
$$=\dfrac{9.95-10}{-4}=0.0125 s$$

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Motion in A Straight Line Test - 47

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Motion in A Straight Line Test - 47

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Question 1

$$\dfrac{n^2}{2n-1}$$

$$\dfrac{2n^2}{n-1}$$

Infinite

n

Solution

Question 2

Both inertial as well as non- inertial frame of reference.

A frame moving with constant velocity w.r.t. an inertial frame.

All reference frames which are at rest w.r.t. an inertial frame.

The reference frame attached to the earth.

Solution

Question 3

$$14.36\ s$$

$$16.18\ s$$

$$10\ s$$

$$16.54\ s$$

Solution

Question 4

$$14.14s$$

$$15.25s$$

$$12.36s$$

$$17.36s$$

Solution

Question 5

$$3\ s$$

$$4.5\ s$$

$$5\ s$$

$$6\ s$$

Solution

Question 6

$$650\ m$$

$$527\ m$$

$$675\ m$$

$$573\ m$$

Solution

Question 7

150 m

145 m

155 m

55 m

Solution

Question 8

$$0.67m/{s^2}$$

$$1m/{s^2}$$

$$1.25m/{s^2}$$

$$1.5m/{s^2}$$

Solution

Question 9

$$216 Km$$

$$512 Km$$

$$728 Km$$

$$1296 Km$$

Solution

Question 10

$$(2s)$$

$$(0.25s)$$

$$(3s)$$

$$(0.0125s)$$

Solution

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