Motion in A Straight Line Test

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Motion in A Straight Line Test - 50

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Question 1
1 / -0

Two particle P and Q are initially $$40m$$ apart P behind Q . Particle P starts moving with a uniform velocity $$10m/s$$ towards Q . Particle Q starting from rest has an acceleration $$2ms^2$$ in the direction of velocity of P. Then the minimum distance between P and Q will be

A
$$45m$$

B
$$15m$$

C
$$35m$$

D
$$30m$$

Solution
Question 2
1 / -0

A man throws balls with the same speed vertically upwards one after the other at an interval of $$2$$ second. What should be the speed of the throw so that more than two balls are in sky at any time? (Given $$g=9.8$$ $${m/s^2}$$)

A
More than $$19.6$$ $${m/s}$$

B
At least $$9.8$$ $${m/s}$$

C
Any speed less than $$19.6$$ $${m/s}$$

D
Only with speed $$19.6$$ $${m/s}$$

Solution
Question 3
1 / -0

A particle moves along a straight line with a uniform acceleration $$-5m/s^{2}$$ and initial velocity $$12.25/s$$ distance traveled by it in $$3rd$$ second

A
0.6725m

B
zero

C
0.25m

D
0.75m

Solution

Distance travelled in $$n^{th}$$ second is given by:-
$$x_n=(4+an)-\dfrac {a}{2}$$
for $$n=3, u=12.25\ m/s,\ a=-5\ m/s^2$$
$$x_3=(12.25-3\times 5)+\dfrac {5}{2}=0.25\ m$$
Question 4
1 / -0

A mass of $$3\ kg$$ descending vertically downward support a mass of $$2\ kg$$ by means of a light string passing over a pulley. At the end of $$5\ s$$ the string breaks. How much high from now the $$2\ kg$$ mass will go ? ($$g=9.8\ m/s^{2}$$)

A
$$4.9\ m$$

B
$$9.8\ m$$

C
$$16.9\ m$$

D
$$2.45\ m$$

Solution

Given that,

Mass $${{m}_{1}}=3\,kg$$

Mass $${{m}_{2}}=2\,kg$$

Now, the initial acceleration a of the system

$$ a=\dfrac{3g-2g}{3+2} $$

$$ a=\dfrac{g}{5} $$

At $$5\ sec$$

Now, from equation of motion

$$ v=u+at $$

$$ v=0+\dfrac{g}{5}\times 5 $$

$$ v=9.8\,m/s $$

When the string breaks at $$t= 5\ s$$,

The mass $$2\ kg$$ more under gravity

So, height is

$$ h=\dfrac{{{v}^{2}}}{2g} $$

$$ h=\dfrac{{{\left( 9.8 \right)}^{2}}}{2\times 9.8} $$

$$ h=4.9\,m $$

Hence, the height is $$4.9\ m$$
Question 5
1 / -0

A ball is thrown upwards with a speed of $$50 m/s$$. Find the distance travelled by the ball in last $$2 s$$ of its ascent. (Take $$10 m/s^2$$)

A
$$35 m$$

B
$$10 m$$

C
$$20 m$$

D
$$5 m$$

Solution

Initial velocity $$-50 m/s$$(opposite to the direction of acceleration)
Final velocity$$=10 m/s$$ (completing the ascent)
Time taken $$=\dfrac{0-(-50)}{acceleration}=\dfrac{50}{10}=5s$$
$$S_{th}=$$Initial acceleration +$$\dfrac{1}{2}$$acceleration ($$2\times$$ (time)-1)
$$S_{4th}=-50+\dfrac{1}{2}\times 10(7)=-15 m$$
$$S_{5th}=-50+\dfrac{1}{2}\times 10(9)=-5 m$$
$$S_{4th}+S_{5th}=-20 m$$(Negative as it is opposite to direction of acceleration)
Question 6
1 / -0

A parachutist after bailing out falls 80 m without air resistance. Then the parachute is opened, and he decelerates at a uniform rate of $$2 m/s^2$$. If he reaches the ground with a speed of $$4 m/s$$, then the height at which he bailed out is [$$g = 10 m/s^2$$]

A
$$576 m$$

B
$$676 m$$

C
$$476 m$$

D
$$376 m$$

Solution

Final velocity after falling 80m,
$$V_s^2=0+2\times 10\times 80(v_c=0)$$
$$V_s=\sqrt{1600}$$
$$V_{s1}=40 m/s$$

Final velocity when reaching ground =4 m/s
$$(V_{s2})^2=V_{s1}^2-2ah$$
$$4^2=40^2-4h$$

$$h=\dfrac{(40-4)(40+4)}{4}$$

$$h=9\times 44=396$$

Total height of bailing $$=396+80=476 m $$
Question 7
1 / -0

A train is at rest. It accelerates for a time $$t_1$$ at a uniform rate $$\alpha$$ and the comes to rest under a uniform retardation rate $$\beta$$ in time.$$t_2$$. $$\dfrac{t_1}{t_2}$$ is equal to

A
$$\dfrac{\beta}{\alpha}$$

B
$$\dfrac{\alpha +\beta}{\alpha}$$

C
$$\dfrac{\alpha + \beta}{\beta}$$

D
$$\alpha \beta$$

Solution

Initially train is at rest.
From the 1st equation of motion,
$$v=u+at$$
$$v=0+at$$
$$v=at$$
From the above equation,
$$\alpha t_1=\beta t_2$$
$$\dfrac{t_1}{t_2}=\dfrac{\beta}{\alpha}$$
The correct option is A.
Question 8
1 / -0

If a body starts from rest,the time in which it covers a particular displacement with uniform acceleration

A
inversely proportional to the square root of the displacement

B
inversely proportional to the displacement

C
diretily proportional to the displacement

D
directly proportional to the square root of the displacement

Solution

Apply equation of kinematic

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

$$ {{t}^{2}}+\dfrac{2ut}{a}-\dfrac{2s}{a}=0 $$

$$ {{t}^{2}}+pt-ks=0 $$

Where, $$p=\dfrac{2u}{a}\,\,and\,\,k=\dfrac{2}{a}$$

$$t=\dfrac{-p\pm \sqrt{{{p}^{2}}+ks}}{2}$$

Hence, time is directly proportional to root of displacement.
Question 9
1 / -0

A stone is thrown vertically upwards. On its way, passes a point P with a speed $$v$$ and point Q is $$20 m$$ higher than Q with a speed $$\cfrac{\sqrt{5}}{3}v$$. The value of speed: [$$g = 10 m/s^2$$]

A
$$20\sqrt5 m/s$$

B
$$15\sqrt5 m/s$$

C
$$ 25 m/s$$

D
$$30 m/s$$

Solution

The velocity at P is V
Heigth difference is 20m
Velocity at Q $$=\dfrac{\sqrt4 v}{3}$$
$$(\dfrac{\sqrt5 v}{3})^2=v^2-2\times 20\times 10$$
$$2\times 20\times 10=v^2-\dfrac{5}{9}v^2$$
$$\dfrac{4}{9}v^2=2\times 20\times 10$$
$$\dfrac{2}{3}v=2\times 10$$
$$v=\dfrac{2\times 10\times 3}{2}=30 m/s$$
Question 10
1 / -0

A body moving with uniform retardation covers $$3\ km$$ before its speed is reduced to half of its initial value. It comes to rest in another distance of :

A
$$1\ km$$

B
$$2\ km$$

C
$$3\ km$$

D
$$\dfrac {1}{2} km$$

Solution

Lets $$u=$$ initial velocity

$$s=3km=3000m$$

$$v=\dfrac{u}{2}$$

3rd equation of motion,

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

$$(\dfrac{u}{2})^2-u^2=2a\times 3000$$

$$-\dfrac{3}{4}u^2=3000\times 2a$$

$$u^2=-8000a$$. . . . . .(1)

Now body comes to rest. 3rd equation of motion is

$$0-(\dfrac{u}{2})^2=2as'$$

$$\dfrac{-u^2}{4}=2as'$$

$$-\dfrac{-8000a}{4}=2as'$$

$$s'=1km$$

The correct option is A.