Motion in a Straight Line — Page 3 | JEE Physics

Q21

In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed '

\upsilon

' more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then '

\upsilon

' is equal to :

A

{{2{a_1}{a_2}} \over {{a_1} + {a_2}}}t

B

\sqrt {2{a_1}{a_2}} t

C

\sqrt {{a_1}{a_2}} t

D

{{{a_1} + {a_2}} \over 2}t

Correct Answer

Option C

Solution

For both car initial speed ( $\mu$ ) = 0 Let the acceleration of car A and car B is

a

1 and

a

2 respectively.

Also let the time taken to reach the finishing point for car A is t1 and for car B is t2.

Let at finishing point speed of car A is

v

1 and speed of car B is

v

2 According to the question, t2 $-$ t1 = t and

v

1 $-$

v

2 =

v

$\Rightarrow$

a

1t1 $-$

a

2t2 =

v

$\Rightarrow$

a

1t1 $-$

a

2(t + t1) =

v

. . . . . . .(1) As, Total distance covered by both car is equal. So, xA = xB $\Rightarrow$

{1 \over 2}{a_1}t_1^2 = {1 \over 2}{a_2}t_2^2

$\Rightarrow$

a

1t

_1^2

=

a

2 (t + t1)2 $\Rightarrow$

\sqrt {{a_1}} .{t_1}

=

\sqrt {{a_2}}

. (t + t1) $\Rightarrow$

\sqrt {{a_1}} .{t_1} - \sqrt {{a_2}} .{t_1} = \sqrt {{a_2}} .t

$\Rightarrow$ t1 =

{{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }}\,\,\,\,\,\,\,.....(2)

Now put the value of t1 in equation (2), (

a

1 $-$

a

2) t1 $-$

a

2t =

v

$\Rightarrow$ (a1 $-$ a2) .

{{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }} - {a_2}t = v

$\Rightarrow$

\left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\sqrt {{a_2}} .t - {a_2}t = v

$\Rightarrow$

\sqrt {{a_1}{a_2}} .t + {a_2}.t - {a_2}t = v

$\Rightarrow$

v

=

\sqrt {{a_1}{a_2}} .t

Q22

A parachutist after bailing out falls

50

m

without friction. When parachute opens, it decelerates at

2\,\,m/{s^2}.

He reaches the ground with a speed of

3

m/s

. At what height, did he bail out?

A

182

m

B

91

m

C

111

m

D

293

m

Correct Answer

Option D

Solution

The velocity of parachutist when parachute opens at 50 m is

u = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 50} = \sqrt {980}

The velocity at ground,

v=3m/s

$\therefore$

S = {{{v^2} - {u^2}} \over {2 \times \left( { - 2} \right)}} = {{{3^2} - 980} \over { - 4}} \approx 243\,m

Initially he has fallen

50

m.

$\therefore$ Total height from where He bailed out

=243+50=293m

Q23

A particle is moving with speed v = b

\sqrt x

along positive x-axis. Calculate the speed of the particle at time t =

\tau

(assume that the particle is at origin t = 0)

A

{{{b^2}\tau } \over {\sqrt 2 }}

B

{{b^2}\tau }

C

{{{b^2}\tau } \over 2}

D

{{{b^2}\tau } \over 4}

Correct Answer

Option C

Solution

v = b

\sqrt x

$\Rightarrow$

{{dx} \over {dt}}

= b

\sqrt x

$\Rightarrow$

\int\limits_0^x {{{dx} \over {\sqrt x }}} = \int\limits_0^t {bdt}

$\Rightarrow$

\left[ {{{{x^{ - {1 \over 2} + 1}}} \over { - {1 \over 2} + 1}}} \right]_0^x

=

b\left[ t \right]_0^t

$\Rightarrow$

{x^{{1 \over 2}}} = {{bt} \over 2}

$\Rightarrow$

x = {{{b^2}{t^2}} \over 4}

$\therefore$ v =

{{dx} \over {dt}}

=

{{{b^2}} \over 4}

$\times$ 2t =

{{{b^2}t} \over 2}

When t = $\tau$ then speed v

= {{{b^2}\tau } \over 2}

Q24

A person travels

x

distance with velocity

v_{1}

and then

x

distance with velocity

v_{2}

in the same direction. The average velocity of the person is

\mathrm{v}

, then the relation between

v, v_{1}

and

v_{2}

will be.

A

\mathbf{V}=\mathbf{V}_{1}+\mathbf{V}_{2}

B

V=\dfrac{v_{1}+V_{2}}{2}

C

\dfrac{1}{\mathrm{v}}=\dfrac{1}{\mathrm{v}_{1}}+\dfrac{1}{\mathrm{v}_{2}}

D

\dfrac{2}{\mathrm{~V}}=\dfrac{1}{\mathrm{v}_{1}}+\dfrac{1}{\mathrm{v}_{2}}

Correct Answer

Option D

Solution

The average velocity is defined as the total displacement divided by the total time.

Here, the person travels the same distance $x$ twice, once with velocity $v_1$ and once with velocity $v_2$ .

The time to travel distance $x$ with velocity $v_1$ is $t_1 = \dfrac{x}{v_1}$ , and the time to travel distance $x$ with velocity $v_2$ is $t_2 = \dfrac{x}{v_2}$ .

The total time is then $t = t_1 + t_2 = \dfrac{x}{v_1} + \dfrac{x}{v_2}$ .

The total displacement is $2x$ .

So, the average velocity $v$ is given by $v = \dfrac{\text{total displacement}}{\text{total time}} = \dfrac{2x}{\dfrac{x}{v_1} + \dfrac{x}{v_2}} = \dfrac{2}{\dfrac{1}{v_1} + \dfrac{1}{v_2}}$ Multiplying both sides by $2$ , we get $\dfrac{2}{v} = \dfrac{1}{v_1} + \dfrac{1}{v_2}$

Q25

The relation between time '

t

' and distance '

x

' is

t=\alpha x^2+\beta x

, where

\alpha

and

\beta

are constants. The relation between acceleration

(a)

and velocity

(v)

is :

A

a=-5 \alpha v^5

B

a=-3 \alpha v^2

C

a=-2 \alpha v^3

D

a=-4 \alpha v^4

Correct Answer

Option C

Solution

The relationship between time (

t

) and distance (

x

) is given by

t = \alpha x^2 + \beta x

, where $\alpha$ and $\beta$ are constants. To find the relation between acceleration (

a

) and velocity (

v

), we can follow these steps: First, we differentiate the given equation with respect to time:

\begin{aligned} & t = \alpha x^2 + \beta x \quad \text{(differentiating with respect to time)} \\ & \frac{dt}{dx} = 2\alpha x + \beta \\ & \frac{1}{v} = 2\alpha x + \beta \\ \end{aligned}

Next, we differentiate again with respect to time to find the acceleration:

\begin{aligned} & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha \frac{dx}{dt} \\ & \text{Since} \quad \frac{dx}{dt} = v, \quad \text{we have:} \\ & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha v \\ & \frac{dv}{dt} = -2\alpha v^3 \\ & \text{Therefore,} \quad a = -2\alpha v^3 \end{aligned}

Q26

A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for time t2 and comes to rest. The correct value of

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

wil be :

A

{{{a_1} + {a_2}} \over {{a_2}}}

B

{{{a_1} + {a_2}} \over {{a_1}}}

C

{{{a_2}} \over {{a_1}}}

D

{{{a_1}} \over {{a_2}}}

Correct Answer

Option C

Solution

From given information : For 1st interval

{a_1} = {{{v_0}} \over {{t_1}}}

v0 = a1 t1 ....... (1) For 2nd interval

{a_2} = {{{v_0}} \over {{t_2}}}

v0 = a2 t2 ..... (2) from (1) & (2) a1 t1 = a2 t2

{{{t_1}} \over {{t_2}}} = {{{a_2}} \over {{a_1}}}

Q27

A body travels

102.5 \mathrm{~m}

in

\mathrm{n}^{\text{th }}

second and

115.0 \mathrm{~m}

in

(\mathrm{n}+2)^{\text{th }}

second. The acceleration is :

A

6.25 \mathrm{~m} / \mathrm{s}^2

B

5 \mathrm{~m} / \mathrm{s}^2

C

12.5 \mathrm{~m} / \mathrm{s}^2

D

9 \mathrm{~m} / \mathrm{s}^2

Correct Answer

Option A

Solution

The distance covered by a body in the

n^{\text{th}}

second can be found using the equation:

S_{n} = u + \dfrac{1}{2}a(2n-1)

where,

S_{n}

is the distance covered in the

n^{\text{th}}

second,

u

is the initial velocity,

a

is the acceleration, and

n

is the nth second. The distance covered in the

n^{\text{th}}

second is given as

102.5 \, \text{m}

, so we have:

102.5 = u + \dfrac{1}{2}a(2n-1)

---- (1) For the

(n + 2)^{\text{th}}

second, the distance covered is:

S_{n+2} = u + \dfrac{1}{2}a(2(n+2)-1)

Substituting

n + 2

in place of

n

, we get:

115.0 = u + \dfrac{1}{2}a(2n+3)

---- (2) Subtracting equation (1) from equation (2), we get:

115.0 - 102.5 = \dfrac{1}{2}a(2n + 3 - 2n + 1)

12.5 = \dfrac{1}{2}a(4)

So, solving for

a

gives:

a = \dfrac{12.5 \times 2}{4} = \dfrac{25}{4} = 6.25 \, \text{m/s}^2

Therefore, the acceleration of the body is:

6.25 \, \text{m/s}^2

Which corresponds to Option A:

6.25 \, \text{m/s}^2

.

Q28

The velocity of a particle is v = v0 + gt + Ft2. Its position is x = 0 at t = 0; then its displacement after time (t = 1) is :

A v0 + g + f

B v0 +

{g \over 2}

+

{F \over 3}

C v0 + 2g + 3F

D v0 +

{g \over 2}

+ F

Correct Answer

Option B

Solution

The velocity of a particle is given by $v = v_0 + gt + Ft^2$ .

Its position is $x = 0$ at $t = 0$ .

To find its displacement after time $t = 1$ , follow these steps: Given: $v = v_0 + gt + Ft^2$ We know that: $\dfrac{dx}{dt} = v_0 + gt + Ft^2$ To find the displacement, integrate both sides with respect to $t$ : $\int_{x = 0}^{x} dx = \int_{t = 0}^{t = 1} (v_0 + gt + Ft^2) \, dt$ This simplifies to: $x = \left[ v_0 t + \dfrac{gt^2}{2} + \dfrac{Ft^3}{3} \right]_{t = 0}^{t = 1}$ Evaluating the integral from $t = 0$ to $t = 1$ : $x = v_0 + \dfrac{g}{2} + \dfrac{F}{3}$ Therefore, the displacement after time $t = 1$ is: $x = v_0 + \dfrac{g}{2} + \dfrac{F}{3}$

Q29

A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when : (i) they are moving in the same direction , and (ii) in the opposite direction is :

A

{{25} \over {11}}

B

{3 \over 2}

C

{{11} \over 5}

D

{5 \over 2}

Correct Answer

Option C

Solution

The total distance to be travelled by the train is 60 + 120 = 180 m.

When the trains are moving in the same direction, relative velocity is v1 – v2 = 80 – 30 = 50 km hr–1 So time taken to cross each other, t1 =

{{180} \over {50 \times {{{{10}^3}} \over {3600}}}}

When the trains are moving in opposite direction, relative velocity is |v1 – (–v2 )| = 80 + 30 = 110 km hr–1 $\therefore$ Time taken to cross each other t2 =

{{180} \over {110 \times {{{{10}^3}} \over {3600}}}}

$\therefore$

{{{t_1}} \over {{t_2}}} = {{{{180} \over {50 \times {{{{10}^3}} \over {3600}}}}} \over {{{180} \over {110 \times {{{{10}^3}} \over {3600}}}}}}

=

{{11} \over 5}

Q30

A car accelerates from rest at a constant rate

\alpha

for some time after which it decelerates at a constant rate

\beta

to come to rest. If the total time elapsed is t seconds, the total distance travelled is :

A

{{4\alpha \beta } \over {(\alpha + \beta )}}{t^2}

B

{{2\alpha \beta } \over {(\alpha + \beta )}}{t^2}

C

{{\alpha \beta } \over {2(\alpha + \beta )}}{t^2}

D

{{\alpha \beta } \over {4(\alpha + \beta )}}{t^2}

Correct Answer

Option C

Solution

{t_1} + {t_2} = t,

V' = 0 + \alpha {t_1}

V = u + at

0 = \alpha {t_1} - \beta {t_2}

{t_2} = {\alpha \over \beta }{t_1} = t

{t_1} = \left( {{\beta \over {\alpha + \beta }}} \right)t

Distance

= {1 \over 2}({t_1} + {t_2}) \times \alpha {t_1}

(area of triangle)

= {1 \over 2}t \times \alpha \left( {{\beta \over {\alpha + \beta }}} \right)t

= {{\alpha \beta } \over {2(\alpha + \beta )}}{t^2}