Motion in A Straight Line Test

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Motion in A Straight Line Test - 28

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Question 1
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Which one of the following represents the displacement time graph of two objects $$A$$ and $$B$$ moving with zero relative speed?

A

B

C

D

Solution

If relative speed $$= 0$$ then velocity of $$A =$$ velocity of $$B$$.
So displacement time graphs of A and B must have same slope (other than zero).
Question 2
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Directions For Questions

A bus starts moving with acceleration 2 $$ms^{-2}$$. A cyclist 96 m behind the bus starts simultaneously towards the bus at a constant speed of 20 m/s.

...view full instructions

After what time will he be able to overtake the bus?

A
$$4 sec$$

B
$$8 sec$$

C
$$12 sec$$

D
$$16 sec$$

Solution

Let after a time $$t$$, the cyclist overtake the bus. Then

$$96+\displaystyle\dfrac { 1 }{ 2 } \times 2\times { t }\displaystyle^{ 2 }=20\times t$$ or $${ t }\displaystyle^{ 2 }-20t+96=0$$

$$\therefore \quad t=\displaystyle\dfrac { 20\pm \sqrt { 400-4\times 96 } }{ 2\times 1 }$$

$$\quad \quad =\displaystyle\dfrac { 20\pm 4 }{ 2 } =8 sec$$ and $$12 sec$$

But since they won't meet again after they have crossed each other once, so only $$8$$ seconds is the answer since they are headed in different directions.
Question 3
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A particle accelerates from rest at a constant rate for some time and attains a velocity of $$8 { m }/{ sec }$$. Afterwards it decelerates with the constant rate and comes to rest. If the total time taken is $$4 sec$$, the distance travelled is

A
$$32 m$$

B
$$16 m$$

C
$$4 m$$

D
None of the above

Solution

Using $$ v=u+at$$
During acceleration,$$8=a{ t }_{ 1 } or { t }_{ 1 }=\displaystyle\dfrac { 8 }{ a }$$ and during deacceleration $$ 0=8-a\left( 4-{ t }_{ 1 } \right)$$ or $${ t }_{ 1 }=\displaystyle\dfrac { 8 }{ a }$$
$$\therefore \quad 8=a\left( 4-\displaystyle\dfrac { 8 }{ a } \right)$$
$$8=4a-8$$ or $$a=4$$ and $${ t }_{ 1 }={ 8 }/{ 4 }=2 sec$$
Now,Distance covered during acceleration $${ s }_{ 1 }=0\times 2+\displaystyle\frac { 1 }{ 2 } \times 4{ \left( 2

\right) }\displaystyle^{ 2 }$$ or $${ s }_{ 1 }=8 m$$
Distance covered during deacceleration $${ s }_{ 2

}=8\times 2-\displaystyle\dfrac { 1 }{ 2 } \times 4\times { \left( 2

\right) }\displaystyle^{ 2 }$$ or $${ s }_{ 2 }=8 m$$
Total distance $$=\quad { s }_{ 1 }+{ s }_{ 2 }=16 m$$
Question 4
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A particle experiences constant acceleration for $$20$$ seconds after starting from rest. If it travels a distance $${ s }_{ 1 }$$ in the first $$10$$ seconds and distance $${ s }_{ 2 }$$ in the next 10 seconds, then

A
$${ s }_{ 2 }={ s }_{ 1 }$$

B
$${ s }_{ 2 }=2{ s }_{ 1 }$$

C
$${ s }_{ 2 }=3{ s }_{ 1 }$$

D
$${ s }_{ 2 }=4{ s }_{ 1 }$$

Solution

Let $$a$$ be the constant acceleration of the particle. Then$$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$$ or $${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times { \left( 10 \right) }\displaystyle^{ 2 }=50a$$and $${ s }_{ 2 }=\left[ 0+\displaystyle\dfrac { 1 }{ 2 } a{ \left( 20 \right) }^{ 2 } \right] -50a=150a$$
$$\therefore \quad { s }_{ 2 }=3{ s }_{ 1 }$$

Alternatively:Let $$a$$ be constant acceleration and

$$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$$, then $${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a$$

Velocity after 10 sec. is $$v=0+10a$$

So, $${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }$$

Let $$a$$ be constant acceleration, using $$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$$

so distance coreved in first 10 seconds $${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a$$

Velocity after 10 sec. is $$v=0+10a$$

So, distance covered in next 10 seconds $${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }$$
Question 5
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Free $$^{238}{U}$$ nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particles and the recoiling nucleus becomes $$x$$, in $$t$$ time after the decay. If a decay takes places, when the train is moving at a uniform speed $$v$$, the distance between the alpha particle and the recoiling nucleus at a time $$t$$ after the decay, as measured by the passenger will be

A
$$x+vt$$

B
$$x-vt$$

C
$$x$$

D
depends on the directions of the train

Solution

Train is moving with constant velocity v.
There won't be any relative velocity change between alpha particle and uranium.
Hence, the distance will still be $$x$$.
Question 6
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A body travels $$2\ m$$ in the first two second and $$2.20\ m$$ in the next $$4$$ second with uniform deceleration. The velocity of the body at the end of $$9$$ second is

A
$$-10\ { m }/{ s }$$

B
$$-0.20\ { m }/{ s }$$

C
$$-0.40\ { m }/{ s }$$

D
$$-0.80\ { m }/{ s }$$

Solution

A to B
$$2=u\times 2+\displaystyle\dfrac { 1 }{ 2 } \times a\times 2\times 2\Rightarrow 1=u+a$$
A to C
$$4.20=u\times 6+\displaystyle\dfrac { 1 }{ 2 } \times a\times 6\times 6\Rightarrow 0.7=u+3a$$

Let the Body travel from $$ A$$ to $$B$$ in $$ 2\ s $$ for a distance of $$ 2\ m $$
Let the Body travel from $$ B$$ to $$C $$ for next $$ 4 s $$ for a distance of $$ 2.20\ m $$
Velocity after $$ 9\ s = ? $$
For $$ A to B $$
WKT, $$ S = ut + \dfrac{1}{2}at^{2} \Rightarrow 2 = 2u + \dfrac{1}{2}a*2*2 \Rightarrow 1 = u + a ... (1)$$
For $$B$$ to $$C$$
WKT, $$ S = ut + \dfrac{1}{2}at^{2} \Rightarrow 4.20 = 6u + \dfrac{1}{2}a*6*6 \Rightarrow 0.7 = u + 3a ... ( 2)$$
From $$(1) $$ and $$( 2) ,$$ we get,
$$ 2a = - 0.3 \Rightarrow a = -0.15 , -ve $$ as it is decreasing acceleration .
$$ u = 1 - a \Rightarrow u = 1 + 0.15 = 1.15 $$
Now, velocity at $$ t = 9 s = v = u + at \Rightarrow v = 1.15 - 0.15*9 = 1.15 - 1.35 = - 0.2 m/s $$
velocity is negative as it is decreasing.
Question 7
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At $$t=0$$, an arrow is fired vertically upwards with a speed of $$98m{ s }^{ -1 }$$. A second arrow is fired vertically upwards with the same speed at $$t=5s$$. Then, (Take $$g=9.8ms^{-2}$$)

A
the arrows will be at the same height above the ground at $$t=12.5s$$.

B
the two arrows will reach back at their starting points at $$t=20s$$ and $$t=25s$$.

C
the ratio of the speeds of the first and the second arrows at $$t=20s$$ will be $$2:1$$.

D
all are correct.

Solution

$$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow 98t-\dfrac { 1 }{ 2 } 9.8{ t }^{ 2 }=98(t-5)-\frac { 1 }{ 2 } 9.8{ (t-5) }^{ 2 }\\ \Rightarrow { t }^{ 2 }-{ (t-5) }^{ 2 }-100=0\\ \Rightarrow 10t=125\\ t=12.5s$$
Now, total time for first arrow will be
$$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow 98t-\dfrac { 1 }{ 2 } 9.8{ t }^{ 2 }=0\\ t(t-20)=0\\ { t }_{ 1 }=20s\\ \Rightarrow { t }_{ 2 }=25s$$
Now at $$t=20s$$
$$v=u+at\\ { v }_{ 1 }=98\quad \& \quad { v }_{ 2 }=98\dfrac { 1 }{ 2 } \\ \dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { 2 }{ 1 } \\ $$
Question 8
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A stone is dropped into a well in which the level of water is $$h$$ below the top of the well. If $$v$$ is velocity of sound, the time $$T$$ after which the splash is heard is given by

A
$$T=\displaystyle\frac { 2h }{ v }$$

B
$$T=\sqrt { \left( \displaystyle\frac { 2h }{ g } \right) } +\displaystyle\frac { h }{ v }$$

C
$$T=\sqrt { \left( \displaystyle\frac { 2h }{ v } \right) } +\displaystyle\frac { h }{ g }$$

D
$$T=\sqrt { \left( \displaystyle\frac { h }{ 2g } \right) } +\displaystyle\frac { 2h }{ v }$$

Solution

Time taken by the stone to reach the water level,
By second kinematic equation,
$$ S = ut + \dfrac{1}{2}gt^{2} \Rightarrow t_{1} = \sqrt{\dfrac{2h}{g}} -------\left ( i \right )$$
Time taken by sound to come to the mouth of the well,
$$ Time = \dfrac{Distance}{Speed} \Rightarrow t_{2} = \dfrac{h}{v} -------\left ( ii \right )$$
$$ \therefore $$ Total time $$ t_{1} + t_{2} = \sqrt{\dfrac{2h}{g}} + \dfrac{h}{v} $$
Question 9
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A balloon starts rising from the ground with an acceleration of $$1.25 { m }/{ { s }^{ 2 } }$$. After $$8 \; s$$, a stone is released from the balloon. The stone will (Taking $$g=10{ m }/{ { s }^{ 2 } }$$)

A
begin to move down after being released

B
reach the ground in $$4$$ $$s$$

C
cover a distance of $$40$$ $$m$$ in reaching the ground

D
will have a displacement of $$50$$ $$m$$

Solution

$$ v=1.25*8{ m }/{ s }=10{ m }/{ s }$$
$$s=\frac { 1 }{ 2 } *1.25*8*8{ m }/{ s=40m }$$

now, $$40=-10t+\dfrac { 1 }{ 2 } *10*{ t }^{ 2 }$$
or, $$5{ t }^{ 2 }-10t-40=0$$
or,$${ t }^{ 2 }-2t-8=0$$

hence, $$t=4s$$

So B is correct.
s$$=$$40m.

displacement$$=$$ 40m.

Just after being released the stone has an upward velocity, so it will move upwards first. So A is wrong.
distance in an upward direction before stopping d.

$$d =\dfrac { { v }^{ 2 }-{ u }^{ 2 } }{ 2g } =\dfrac { { 10 }^{ 2 }-{ 0 }^{ 2 } }{ 2*10 } =5m $$

and distance$$=$$ s+2d$$=$$50m
So the distance covered is 50 m and the displacement is 40 m. So C and D are wrong.
Question 10
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Figure shows the $$V-T$$ graph for two particles $$P$$ and $$Q$$. The relative velocity of $$P$$ w.r.t. $$Q$$ is:

A
is zero.

B
is non-zero but constant

C
continuously decreases

D
continuously increases

Solution

The difference in velocities is increasing with time as both of them have more constant but different acceleration.