Motion in A Straight Line Test

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Motion in A Straight Line Test - 69

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Question 1
1 / -0

Two towns are connected through a straight highway. A car runs between two towns with constant speed $$36\ km/hr$$. One day track of $$100m$$ is to be covered with speed $$18\ km/hr$$. For this car accelerates and retards at rate $$1\ { m/sec }^{ 2 }$$. By what time the car will get delayed that day. (When car reaches at any town its speed is $$36\ km/hr$$.)

A
$$8.5\ s$$

B
$$10.5\ s$$

C
$$12.5\ s$$

D
$$6.5\ s$$

Solution
Question 2
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Particle has initial velocity $$9\ ms^{-1}$$ due east and constant acceleration of $$2\ ms^{-2}$$ due west. If the distance covered by it in fifth second of its motion is $$\dfrac{n}{10}m$$, then the value of $$'n'$$ is:

A
$$5$$

B
$$1$$

C
$$3$$

D
$$2$$

Solution

Here the particle velocity will become zero in 4.5 seconds,as
$$v=u+at$$
$$0=9-2t$$
$$t=4.5sec$$
Displacement in the same period will be
$$v^2-u^2=2as$$
$$-81=2\times (-2)s$$
$$s_{4.5}=20.25m$$
Let us call the point of velocity being 0 as Point A
Now body will start moving backwards,
Displacement in next 0.5 sec from point A,
$$s=ut+\frac{1}{2}at^2$$
$$s_{0.5}=\frac{1}{2}\times(-2)\times 0.25=-0.25m$$
Hence distance covered is 0.25m
(negative sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)
Distance traveled in 4 seconds
$$s_4=36-16=20m$$
Distance covered in 5th second is
$$D=s_{4.5}+s_{0.5}-s_4$$
$$D=20.25+0.25-20$$
$$D=0.5m$$
which means the distance covered by particle in 5th second of its motion is $$\frac{n}{10}=\frac{5}{10}=0.5$$
Therefore value of n is 5
Question 3
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For a particle undergoing rectilinear motion with uniform acceleration , the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude if initial velocity for this time interval. Then the ratio of initial speed to the final speed for this time interval is:

A
$$\sqrt { 2 }$$

B
$$2$$

C
$$\sqrt { 3 }$$

D
$$3$$

Solution
Question 4
1 / -0

A stone thrown upwards with speed $$u$$ attains maximum height $$h$$. Another stone thrown upwards from the same point with sapped $$2u$$ attains maximum height $$H$$. What is the relation between $$h$$ and $$H$$?

A
$$2h=H$$

B
$$3h=H$$

C
$$4h=H$$

D
$$5h=H$$

Solution

$$h=\dfrac{u^2}{2g}$$
$$H=\dfrac{(2u)^2}{2g}$$
$$H=4h$$
Question 5
1 / -0

Two trains one of length $$100\ m$$ and another of length $$125m$$, are moving in mutually opposite directions along parallel lines, meet each other, each with speed $$10m/s$$. If their acceleration are $$0.3m/{s}^{2}$$ and $$0.2m/{s}^{2}$$ respectively, then the time they take to pass each other will be

A
$$5\ s$$

B
$$10\ s$$

C
$$15\ s$$

D
$$20\ s$$

Solution

$$\textbf{Step 1: Drawing initial and final position [Refer Fig. 1 and 2]}$$

$$\textbf{Step 2: Solving in frame of reference of Train 2.}$$

Initial speed $$u_{{1}/{2}} = 10 - (-10)m/s$$   $$= 20\ m/s$$

Acceleration $$a_{\frac {1}{2}} = 0.3 - (-0.2) m/s^{2}$$    $$= 0.5\ m/s^{2}$$
Displacement $$s = 100 - (-125) m$$   $$= 225\ m$$

$$\textbf{Step 3: Apply equation of motion}$$
Since acceleration is constant, therefore applying equation of motion
$$s = ut + \dfrac {1}{2} at^{2}$$

$$\Rightarrow$$ $$225 = 20t + \dfrac {1}{2} (0.5) t^{2}$$

  $$\Rightarrow$$ $$t^{2} = 80 t - 900 = 0$$

Solving the quadratic eqn
$${t = 10s}$$
Hence option $$B$$ is correct.
Question 6
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Two guns are mounted (fixed) on two vertical cliffs that are very high from the ground as shown in figure. The muzzle velocity of the shell from $${G}{1}$$ is $${u}{1}$$ and that from $${G}{2}$$ is $${u}{2}$$. The guns aim exactly towards each other The ratio $${u}{1}:{u}{2}$$ such that the shells collide with each other in air is (Assume that there is no resistance of air)

A
$$1:2$$

B
$$1:4$$

C
$$1:6$$

D
will not collide for any ratio

Solution
Question 7
1 / -0

A car starting from rest, is accelerated at a constant rate $$ \alpha $$ until it attains a speed $$V$$ . It is then retarded at a constant rate $$\beta $$ until it comes to rest. The average speed of the car during its entire journey is :

A
Zero

B
$$ \dfrac{\alpha v }{2} $$

C
$$\dfrac {\beta v }{2} $$

D
$$ \dfrac{v }{2} $$

Solution

The average speed is given as,

$${v_{avg}} = \dfrac{s}{t}$$

$${v_{avg}} = \dfrac{{\dfrac{{{V^2}}}{{2\alpha }} + \dfrac{{{V^2}}}{{2\beta }}}}{{\dfrac{V}{\alpha } + \dfrac{V}{\beta }}}$$

$${v_{avg}} = \dfrac{V}{2}$$
Question 8
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A particle starts from rest and travels a total distance of 18m along a straight path.The first half of the distance was travelled with a uniform acceleration of 2 $$ms^{-2}$$ and the rest uniform velocity.The average velocity for the whole journey is (in $$ms^{-1}$$)

A
3

B
4

C
6

D
9

Solution
Question 9
1 / -0

A barometer kept in an elevator reads $$76 cm $$ when it is at rest. What will be barometric reading when the elevator accelerates upwards?

A
$$ \dfrac {(g+a)h_0}{g} $$

B
$$ \dfrac {gh_0}{g+a} $$

C
$$ \dfrac {a}{g} h_0 $$

D
$$ \dfrac{g}{a} h_0 $$

Solution

When a body moves upward the acceleration due to gravity is,
$${g'} = g + a$$

Hence, the pressure is

$$P = h_0\rho \left( {g + a} \right)$$

$$ = \dfrac{{h_0\left( {g + a} \right)}}{g}$$

Hence, the barometer reading will be more than the measured value at rest.
Question 10
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A machine gun is mounted on a $$2000 kg $$ vehicle on a horizontal smooth road (friction negligible). The gun fires $$10 $$ bullets per sec with a velocity of $$500 m/s $$ . If the mass of each bullet be $$10 g $$ , what is the acceleration produced in the vehicle?

A
$$ 25 cm/s^2 $$

B
$$ 0.025 m/s^2 $$

C
$$ 0.50 cm/s^2 $$

D
$$ 50 m/s^2 $$

Solution

The momentum of each bullet is given as,

$${P_b} = {m_b}{v_b}$$

$$ = 10 \times {10^{ - 3}} \times 500$$

$$ = 5\;{\rm{kgm/s}}$$

The force on the vehicle will be equal to change in momentum in $$1\;{\rm{sec}}$$.

The force is given as,

$$F = n{P_b}$$

$$ = 10 \times 5$$

$$ = 50\;{\rm{N}}$$

The acceleration of the vehicle is given as,

$$a = \dfrac{F}{{{m_v}}}$$

$$ = \dfrac{{50}}{{2000}}$$

$$ = 0.025\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$

Thus, the acceleration produced in the vehicle is $$0.025\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$.