Motion in a Straight Line Questions — JEE Physics

Q1

A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is :

A 50 m

B 25 m

C 45 m

D 35 m

Correct Answer

Option C

Solution

For particle (1)

20 + h = 10t + {1 \over 2}g{t^2}

...... (i) For particle (2)

h = {1 \over 2}g{t^2}

..... (ii) put equation (ii) in equation (i)

20 + {1 \over 2}g{t^2} = 10t + {1 \over 2}g{t^2}

t = 2 sec. Put in equation (ii)

h = {1 \over 2}g{t^2}

= {1 \over 2} \times 10 \times {2^2}

h = 20 m The height of the building

= 25 + 20

= 45 m

Q2

The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is :

A 2.0

B 1.0

C 0.5

D 1.5

Correct Answer

Option C

Solution

S = 4 cm

v{'_4} = {v \over 3}

, a = constant

{v_{4 + x}} = 0

\left( {{v^2} - {{{v^2}} \over a}} \right) = 2a(4)

({v^2} - 0) = 2a(4 + x)

{4 \over {4 + x}} = {8 \over 9}

\Rightarrow x = 0.5

m

Q3

The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

A

a + {{{b^2}} \over {c}}

B

a + {{{b^2}} \over {4c}}

C

a + {{{b^2}} \over {3c}}

D

a + {{{b^2}} \over {2c}}

Correct Answer

Option C

Solution

x = at + bt2 – ct3

V = {{dx} \over {dt}} = a + 2bt - 3c{t^2}

a = {{dv} \over {dt}} = 2b - 6ct

Put acceleration = 0

\Rightarrow t = {b \over {3c}}

Now V at t =

{b \over {3c}}

V = a + {{{b^2}} \over {3c}}

Q4

An automobile, travelling at

40\,

km/h, can be stopped at a distance of

40\,

m by applying brakes. If the same automobile is travelling at

80\,

km/h, the minimum stopping distance, in metres, is (assume no skidding) :

A

45\,

m

B

100\,

m

C

150\,

m

D

160\,

m

Correct Answer

Option D

Solution

An automobile traveling at 40 km/h can be stopped within a distance of 40 meters by using brakes.

If the same automobile is traveling at 80 km/h, we need to determine the minimum stopping distance in meters, assuming no skidding occurs.

First Case Initial speed, $u_1 = 40 \, \text{km/h}$ Final speed, $v_1 = 0 \, \text{m/s}$ Stopping distance, $s_1 = 40 \, \text{m}$ Using the equation: $v^2 - u^2 = 2as$ For the first case: $0^2 - 40^2 = 2a \times 40$ $-1600 = 80a$ $a = -20 \, \text{m/s}^2$ Second Case Initial speed, $u_2 = 80 \, \text{km/h}$ Final speed, $v_2 = 0 \, \text{m/s}$ Similarly, using the same equation: $0^2 - 80^2 = 2a s_2$ $-6400 = 2a s_2$ Since $a = -20 \, \text{m/s}^2$ , divide both sides of the equation for the second case by the first case: $\dfrac{s_2}{40} = \dfrac{80^2}{40^2}$ $s_2 = \dfrac{80 \times 80}{40}$ $s_2 = 160 \, \text{m}$ Thus, the minimum stopping distance when the automobile is traveling at 80 km/h is 160 meters.

Q5

An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :

A

{{u + v} \over 2}

B

\sqrt {{{{v^2} - {u^2}} \over 2}}

C

{{v - u} \over 2}

D

\sqrt {{{{v^2} + {u^2}} \over 2}}

Correct Answer

Option D

Solution

Let initial speed of train u. When midpoint of the train reach the signal post it's velocity becomes v0. $\therefore$

v_0^2 = {u^2} + 2as

.......(1) When train passes the signal post completely it's velocity becomes v. $\therefore$

{v^2} = v_0^2 + 2as

......(2) Subtracting (2) from (1) we get,

v_0^2 - {v^2} = {u^2} - v_0^2

\Rightarrow v_0^2 + v_0^2 = {u^2} + {v^2}

\Rightarrow {v_0} = \sqrt {{{{u^2} + {v^2}} \over 2}}

Q6

A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t1. If he remains stationary on a moving escalator then the escalator takes him up in time t2. The time taken by him to walk up on the moving escalator will be :

A

{{{t_1}{t_2}} \over {{t_2} - {t_1}}}

B

{{{t_1} + {t_2}} \over 2}

C

{{{t_1}{t_2}} \over {{t_2} + {t_1}}}

D

{t_2} - {t_1}

Correct Answer

Option C

Solution

L = Length of escalator

{V_{b/esc}} = {L \over {{t_1}}}

When only escalator is moving.

{V_{esc}} = {L \over {{t_2}}}

when both are moving

{V_{b/g}} = {V_{b/esc}} + {V_{esc}}

{V_{b/g}} = {L \over {{t_1}}} + {L \over {{t_2}}} \Rightarrow \left[ {t = {L \over {{V_{b/g}}}} = {{{t_1}{t_2}} \over {{t_1} + {t_2}}}} \right]

Q7

A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around : (takes the value of g as 10 m/s2)

A 300 m

B 200 m

C 125 m

D 250 m

Correct Answer

Option C

Solution

Object is projected as shown so as per motion under gravity

S = ut + {1 \over 2}a{t^2}

- 75 = + 10t + {1 \over 2}( - 10){t^2} \Rightarrow t = 5

sec Object takes t = 5 s to fall on ground Height of balloon from ground H = 75 + ut = 75 + 10 $\times$ 5 = 125 m

Q8

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching

{h \over 3}

in both the directions.

A

{{\sqrt 2 - 1} \over {\sqrt 2 + 1}}

B

{1 \over 3}

C

{{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}

D

{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}

Correct Answer

Option C

Solution

u = \sqrt {2gh}

Now,

S = {h \over 3}

a = $-$ g

S = ut + {1 \over 2}a{t^2}

{h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}

{t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0

From quadratic equation

{t_1},{t_2} = {{\sqrt {2gh} \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}

{{{t_1}} \over {{t_2}}} = {{\sqrt {2gh} - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh} + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}

Q9

The instantaneous velocity of a particle moving in a straight line is given as

V = \alpha t + \beta {t^2}

, where

\alpha

and

\beta

are constants. The distance travelled by the particle between 1s and 2s is :

A 3

\alpha

+ 7

\beta

B

{3 \over 2}\alpha + {7 \over 3}\beta

C

{\alpha \over 2} + {\beta \over 3}

D

{3 \over 2}\alpha + {7 \over 2}\beta

Correct Answer

Option B

Solution

V = \alpha t + \beta {t^2}

{{ds} \over {dt}} = \alpha t + \beta {t^2}

\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} }

{S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2

As particle is not changing direction So distance = displacement Distance =

\left[ {{{\alpha [4 - 1]} \over 2} + {{\beta [8 - 1]} \over 3}} \right]

= {{3\alpha } \over 2} + {{7\beta } \over 3}

Q10

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap ? (Take g = 9.8 m/s2)

A 3 drops / 2 sconds

B 2 drops / second

C 1 drop / second

D 1 drop / 7 seconds

Correct Answer

Option C

Solution

In 4 sec. 1st drop will travel

\Rightarrow {1 \over 2}

$\times$ (9.8) $\times$ (4)2 = 78.4 m $\therefore$ 2nd drop would have travelled $\Rightarrow$ 78.4 $-$ 34.3 = 44.1 m.

Time for 2nd drop

\Rightarrow {1 \over 2}

(9.8)t2 = 44.1 t = 3 sec $\therefore$ each drop have time gap of 1 sec $\therefore$ 1 drop per sec