Motion in A Straight Line Test

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Motion in A Straight Line Test - 66

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Weekly Quiz Competition

Question 1
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Which of the following statements is correct for a particle travelling with a constant speed?

A
Its position remains constant as time passes.

B
It covers equal distances in unequal time intervals.

C
Its acceleration is zero.

D
It does not change its direction of motion.
Question 2
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A car and a bike start racing in a straight line. The distance of the finish line from the starting line is $$100m$$. The minimum acceleration of the car for it to win, if it accelerates uniformly starting from rest and the bike moves with a constant velocity of $$10m/s$$, is

A
$$0.5{m}/s^{2}$$

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

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

D
$$3{m}/s^{2}$$

Solution

For bike :
Distance cover to finish the race $$d = 100 \ m$$
Velocity   $$v = 10 \ m/s$$
Time taken   $$t = \dfrac{d}{v} = \dfrac{100}{10} = 10 \ s$$
For car :
Let the acceleration be $$a$$.
Initial velocity of car   $$u = 0$$
Using    $$S = ut+\dfrac{1}{2}at^2$$
where $$t = 10 \ s$$
So,    $$100 = 0+\dfrac{1}{2}\times a\times 10^2$$
$$\implies \ a = 2 \ m/s^2$$
Question 3
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A particle starts moving along a line from rest and comes to rest after moving distance $$d$$. During its motion, it had a constant acceleration $$f$$ over $$2/3$$ of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is:

A
$$\sqrt {\dfrac{2d}{3f}}$$

B
$$2\sqrt {\dfrac{d}{3f}}$$

C
$$\sqrt {\dfrac{3d}{f}}$$

D
$$\sqrt {\dfrac{3d}{2f}}$$

Solution

Let velocity at B be $$v$$.
From A to B :
Initial velocity, $$u = 0$$
Acceleration, $$a= f$$
Time taken, $$t=t_1$$
Using Newton's first equation of motion,
$$v=u+at$$
Velocity at B, $$v_1 = 0+ft_1 = ft_1$$
$$v_1= ft_1$$ ....(1)
Using Newton's second equation of motion,
$$s=ut+\dfrac 12 at^2$$
Distance covered, $$s= \dfrac{2d}{3}$$
$$\dfrac{2d}{3} = 0+\dfrac{1}{2}ft_1^2$$
$$\implies \ t_1 = \sqrt{\dfrac{4d}{3f}}$$
$$\implies \ t_1 =2 \sqrt{\dfrac{d}{3f}}$$ .....(b)

From B to C :
Initial velocity, $$u = v_1$$
Final velocity, $$v = 0$$
Acceleration, $$a$$
Time taken, $$t=t_2$$
Using Newton's first equation of motion,
$$v=u+at$$
$$0 = v_1+at_2$$
$$\implies = v = -at_2$$ ....(2)
From (1) and (2), we get
$$a = -\dfrac{ft_1}{t_2}$$
Using Newton's third equation of motion
$$2as=v^2-u^2$$
$$2a\dfrac{d}{3}=0^2-v_1^2$$
$$\dfrac{d}{3} = \dfrac{-v^2}{2a} = \dfrac{-(ft_1)^2}{2\times (-ft_1/t_2)} = \dfrac{ft_1t_2}{2}$$ ....(a)
From (a) and (b), $$\dfrac{d}{3} = \dfrac{ft_2}{2}\times \sqrt{\dfrac{4d}{3f}}$$
$$\implies \ t_2 = \sqrt{\dfrac{d}{3f}}$$
Thus total time taken, $$t = t_1+t_2 = 2 \sqrt{\dfrac{d}{3f}}+\sqrt{\dfrac{d}{3f}} = 3\sqrt { \dfrac{d}{3f}}$$
$$t=\sqrt { \dfrac{3d}{f}}$$
Question 4
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A train of length / = 350 m starts moving rectilinearly with constant acceleration a = $$3.0 \times 10^{-2}$$ $$ms^{-2}$$. After r = 30 s from start, the locomotive headlight is switched on (event 1), and 60 s after this event, the tail signal light is switched on (event 2).
a.Find the distance between these events in the reference frame fixed to the train and to the Earth.
b.How and at what constant velocity v relative to the Earth must a certain reference frame R move for the two events to occur in it at the same point?

A
a. 350 m is with respect to train; 242 m is with respect to Earth; b. 4.03 m/s

B
a. 350 m is with respect to train; 252 m is with respect to Earth; b. 4.03 m/s

C
a. 350 m is with respect to train; 262 m is with respect to Earth; b. 4.03 m/s

D
a. 250 m is with respect to train; 242 m is with respect to Earth; b. 4.03 m/s

Solution
Question 5
1 / -0

A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ and comes to rest. It total time elapsed is $$t$$, then maximum velocity acquired by car will be:

A
$$\cfrac { \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) t }{ \alpha \beta } $$

B
$$\cfrac { \left( { \alpha }^{ 2 }+{ \beta }^{ 2 } \right) t }{ \alpha \beta } $$

C
$$\cfrac { \left( { \alpha }^{ }+{ \beta }^{ } \right) t }{ \alpha \beta } $$

D
$$\cfrac { \left( { \alpha }^{ }{ \beta }^{ } \right) t }{ \alpha +\beta } $$

Solution

$$\textbf{Step 1 - Calculation of acceleration and deacceleration :}$$
Apply first equation of motion,
From rest, car accelerates for time $$t$$,
$$V=u+at=0+\alpha t$$
So $$\alpha = \dfrac {V}{t_{1}}$$ $$....(1)$$
Let maximum velocity is $$V$$.
The car begins to deaccelerate after obtaining maximum velocity
$$0=V-\beta (t-t_{1})$$
$$\therefore \beta = \dfrac {V}{t - t_{1}}$$ $$....(2)$$

$$\textbf{Step 2 - Calculation of velocity :}$$
By adding equation $$(1)$$ and $$(2)$$
$$t = V\left (\dfrac {1}{\alpha} + \dfrac {1}{\beta}\right )$$
$$V = \dfrac {\alpha \beta t}{\alpha + \beta}$$

Hence, the maximum velocity acquired by car is
$$V = \dfrac {\alpha \beta t}{\alpha + \beta}$$
Question 6
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Two cars are travelling towards each other on a straight road at velocities $$15ms^{-1}$$ and $$16ms^{-1}$$ respectively. When they are $$150$$m apart, both the drivers apply the breaks and the cars decelerates at $$3ms^{-2}$$ and $$4ms^{-2}$$ until they stop. How for apart will they be when they have come to a stop.

A
$$86.5$$m

B
$$89.5$$m

C
$$85.5$$m

D
$$81.36$$m

Solution

solving with respect to the car $$1$$ moving with speed $$15m/s$$.
using concept of relative motion we get $$V_{21}=V_{2g}-V_{1g}$$
same with acceleration $$a_{21}=a_{2g}-a_{1g}$$
so velocity car $$2$$ is $$15+16=31m/s, acc=4+3=7m/s^2 $$and $$ separation =150m$$
distance covered before coming to rest $$\frac{31^2}{2\times7}=S=>S=68.64$$
so now separation between them is $$150-68.64=81.36m$$
so none of the option is correct.
Question 7
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A man running with a uniform speed 'u' on a straight road observes a stationary bus at a distance 'd' ahead of him. At that instance , the bus starts with an acceleration 'a' .The condition that he would be able to catch the bus is:

A
$$\displaystyle d \, \leq \, \frac{u^2}{a}$$

B
$$\displaystyle d \, \leq \, \frac{u^2}{2a}$$

C
$$\displaystyle d \, \leq \, \frac{u^2}{3a}$$

D
$$\displaystyle d \, \leq \, \frac{u^2}{4a}$$

Solution

Relative speed of man w.r.t bus $$=u$$
and acceleration w.r.t bus $$=-a$$
if man catch the bus
$$d\leq x$$ where x is calculated as
$$v^{2}-u^{2}=-2ax$$
$$v=$$ final velocity w.r.t bus that is zero.
$$0-u^{2}=-2ax$$
$$x=\frac{u^{2}}{2a}$$
hence $$d \leq \dfrac{u^{2}}{2a}$$
Question 8
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A body is falling freely from a point $$A$$ at a certain height from the ground and passes through points $$B,C$$ and $$D$$ (vertically as shown) so that $$BC=CD$$. The time taken by the particle to move from $$B$$ to $$C$$ is $$2$$ seconds and from $$C$$ to $$D$$ $$1$$ second. Time taken to move from $$A$$ to $$B$$ in seconds is

A
$$0.6$$

B
$$0.5$$

C
$$0.2$$

D
$$0.4$$

Solution

Velocity of the body at point A    $$u_A = 0$$
Let the velocity of body at points B and C be $$u_B$$ and $$u_C$$ respectively.
Let $$BC = CD = S$$
For journey $$A$$ to $$B$$ :
Using   $$u_B =u_A +gt_{AB}$$
where $$t_{AB}$$ is the time taken to move from A to B.
$$\therefore$$   $$u_B = 10t_{AB}$$   .......(1)        $$(\because u_A = 0)$$
For journey $$B$$ to $$C$$ :
Using   $$u_C = u_B +gt_{BC}$$
$$\therefore$$   $$u_C = u_B+20$$
Using   $$S = u_Bt_{BC}+\dfrac{10}{2}t_{BC}^2$$
where $$t_{BC} = 2$$
We get   $$S = 2u_B+5(2)^2 = 2u_A+20$$        .....(2)
For journey $$C$$ to $$D$$ :
Using   $$S = u_Ct_{CD}+\dfrac{10}{2}t_{CD}^2$$
where $$t_{CD} = 1$$
We get   $$S = u_C+5(1)^2 = u_C+5$$        .....(3)
Equation (2) - (1), we get $$0 = 2u_B-u_C + 15$$
$$\implies \ 2u_B = u_C-15$$
Or     $$2u_B = u_B+20-15$$
$$\implies \ u_B = 5$$
From (1), we get   $$5 = 10t_{AB}$$
$$\implies \ t_{AB} = 0.5   \ s$$
Question 9
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Directions For Questions

An elevator without a ceiling is ascending up with an acceleration up with an acceleration of $$5\ ms^{-2}$$. A boy on the elevator shoots a ball in vertically upward direct from a height $$2\ m$$ above the floor of elevator. At this instant, the elevator is moving up with a velocity of $$10\ ms^{-1}$$ and floor, of the elevator is at a height of $$50\ m$$ from the ground. The initial speed of the ball is $$15\ ms^{-1}$$ w.r.t. the elevator. Consider the duration for which the ball strikes the floor of the elevator in answering following question:

...view full instructions

The maximum height reached by the ball as measured from the ground would be:

A
$$52\ m$$

B
$$31.25\ m$$

C
$$83.25\ m$$

D
$$82.56\ m$$

Solution

Initial velocity, $$u=15+10=25$$m/s.
at maximum height, $$V=0$$
$$\therefore$$ $$V=u+at$$
$$0=25-10t$$
$$\Rightarrow$$ $$t=2.5s$$
Time to reach on the floor $$t_0=2.13 \ sec$$
Since $$t<t_0,$$, therefore as per the question ball is at
its maximum height at 2.13s.
We have, $$x={ x }_{ 0 }+ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$$
Taking ground as reference, $${ x }_{ 0 }=50+2=52$$
$$\therefore$$ $${ x }_{ max }=52+25\times 2.5+\dfrac { 1 }{ 2 } \times \left( -10 \right) \times { 2.13 }^{ 2 }$$
$$=82.56m$$
Question 10
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Directions For Questions

An elevator without a ceiling is ascending up with an acceleration up with an acceleration of $$5\ ms^{-2}$$. A boy on the elevator shoots a ball in vertically upward direct from a height $$2\ m$$ above the floor of elevator. At this instant, the elevator is moving up with a velocity of $$10\ ms^{-1}$$ and floor, of the elevator is at a height of $$50\ m$$ from the ground. The initial speed of the ball is $$15\ ms^{-1}$$ w.r.t. the elevator. Consider the duration for which the ball strikes the floor of the elevator in answering following question:

...view full instructions

The maximum separation between the floor of elevator and the ball during its flight is:

A
$$30\ m$$

B
$$15\ m$$

C
$$7.62\ m$$

D
$$9.5\ m$$

Solution

Initial point of floor is considered as reference point.
We have, $$x={ x }_{ 0 }+ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$$
$${ x }_{ b }=2+25t-5{ t }^{ 2 }$$
$${ x }_{ ele }=10t+{ 1/2\times 10t }^{ 2 }=10t+{5t }^{ 2 }$$
Separation, $$\Delta ={ x }_{ b }-{ x }_{ ele }$$
$$\Rightarrow \Delta =2+15t-10{ t }^{ 2 }$$
$$\dfrac { d\Delta }{ dt } =0\Rightarrow 15-20t=0$$
$$\Rightarrow t=\dfrac { 3 }{ 4 } $$
as $$\dfrac { { d }^{ 2 }\Delta }{ { dt }^{ 2 } } <0$$, $$\Delta $$ at $$t=\dfrac { 3 }{ 4 } $$ will be max.
$$\therefore$$ $$\Delta $$ at $$t=\dfrac { 3 }{ 4 } =2+15\times \dfrac { 3 }{ 4 } -10\times { \left( \dfrac { 3 }{ 4 } \right) }^{ 2 }$$
$$=7.625m$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 66

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Motion in A Straight Line Test - 66

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SHARING IS CARING

Question 1

Its position remains constant as time passes.

It covers equal distances in unequal time intervals.

Its acceleration is zero.

It does not change its direction of motion.

Question 2

$$0.5{m}/s^{2}$$

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

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

$$3{m}/s^{2}$$

Solution

Question 3

$$\sqrt {\dfrac{2d}{3f}}$$

$$2\sqrt {\dfrac{d}{3f}}$$

$$\sqrt {\dfrac{3d}{f}}$$

$$\sqrt {\dfrac{3d}{2f}}$$

Solution

Question 4

a. 350 m is with respect to train; 242 m is with respect to Earth; b. 4.03 m/s

a. 350 m is with respect to train; 252 m is with respect to Earth; b. 4.03 m/s

a. 350 m is with respect to train; 262 m is with respect to Earth; b. 4.03 m/s

a. 250 m is with respect to train; 242 m is with respect to Earth; b. 4.03 m/s

Solution

Question 5

$$\cfrac { \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) t }{ \alpha \beta } $$

$$\cfrac { \left( { \alpha }^{ 2 }+{ \beta }^{ 2 } \right) t }{ \alpha \beta } $$

$$\cfrac { \left( { \alpha }^{ }+{ \beta }^{ } \right) t }{ \alpha \beta } $$

$$\cfrac { \left( { \alpha }^{ }{ \beta }^{ } \right) t }{ \alpha +\beta } $$

Solution

Question 6

$$86.5$$m

$$89.5$$m

$$85.5$$m

$$81.36$$m

Solution

Question 7

$$\displaystyle d \, \leq \, \frac{u^2}{a}$$

$$\displaystyle d \, \leq \, \frac{u^2}{2a}$$

$$\displaystyle d \, \leq \, \frac{u^2}{3a}$$

$$\displaystyle d \, \leq \, \frac{u^2}{4a}$$

Solution

Question 8

$$0.6$$

$$0.5$$

$$0.2$$

$$0.4$$

Solution

Question 9

$$52\ m$$

$$31.25\ m$$

$$83.25\ m$$

$$82.56\ m$$

Solution

Question 10

$$30\ m$$

$$15\ m$$

$$7.62\ m$$

$$9.5\ m$$

Solution

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