Motion in a Straight Line — Page 9 | NEET Physics

Q81

The relation between time t and distance x for a moving body is given as t = mx2 + nx, where m and n are constants. The retardation of the motion is : (When v stands for velocity)

A 2 mv3

B 2 mnv3

C 2nv3

D 2n2v3

Correct Answer

Option A

Solution

The relationship between time $t$ and distance $x$ for a moving body is given by $t = mx^2 + nx$ , where $m$ and $n$ are constants.

To determine the retardation (negative acceleration) of the motion, let's follow the steps to derive it: Given: $t = mx^2 + nx$ First, differentiate $t$ with respect to $x$ : $\dfrac{dt}{dx} = 2mx + n$ Since velocity $v$ is defined as $\dfrac{dx}{dt}$ , we can write: $\dfrac{1}{v} = \dfrac{dt}{dx} = 2mx + n$ Thus, $v = \dfrac{1}{2mx + n}$ Next, to find the acceleration $a$ (which is the derivative of velocity with respect to time), we start with the chain rule: $\dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt}$ Since $v = \dfrac{dx}{dt}$ , substituting $x$ gives: $\dfrac{dx}{dt} = v$ Rewrite it: $\dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot v$ Differentiate $v$ with respect to $x$ : $v = (2mx + n)^{-1}$ $\dfrac{dv}{dx} = -\dfrac{2m}{(2mx + n)^2}$ Then, $\dfrac{dv}{dt} = -\dfrac{2m}{(2mx + n)^2} \cdot v$ Substitute $v = \dfrac{1}{2mx + n}$ into the expression: $\dfrac{dv}{dt} = -\dfrac{2m}{(2mx + n)^2} \cdot \dfrac{1}{2mx + n}$ Simplify the expression: $\dfrac{dv}{dt} = -2m \left( \dfrac{1}{2mx + n} \right)^3$ $a = -2m v^3$ So, the retardation (negative acceleration) is: $a = -2m v^3$

Q82

Train A and train B are running on parallel tracks in the opposite directions with speeds of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/ hour. Speed (in ms–1) of this person as observed from train B will be close to : (take the distance between the tracks as negligible)

A 30.5 ms–1

B 29.5 ms–1

C 31.5 ms–1

D 28.5 ms–1

Correct Answer

Option B

Solution

Velocity of man with respect to ground

{\overrightarrow V _{m/g}}

=

{\overrightarrow V _{m/A}}

+

{\overrightarrow V _{A}}

= -1.8 + 36 Velocity of man w.r.t. B

{\overrightarrow V _{m/B}}

=

{\overrightarrow V _{m}}

-

{\overrightarrow V _{B}}

= –1.8 + 36 – (–72) = 106.2 km/hr = 29.5 m/s

Q83

The position vector of a particle changes with time according to the relation

\overrightarrow r (t) = 15{t^2}\widehat i + (4 - 20{t^2})\widehat j

What is the magnitude of the acceleration at t = 1 ?

A 50

B 25

C 40

D 100

Correct Answer

Option A

Solution

\overrightarrow r = \left( {15{t^2}} \right)\widehat i + \left( {4 - 20{t^2}} \right)\widehat j

\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \left( {30t} \right)\widehat i - \left( {40t} \right)\widehat j

\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \left( {30} \right)\widehat i - \left( {40} \right)\widehat j

\left| {\overrightarrow a } \right| = 50

Q84

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

A 60 m

B 40 m

C 20 m

D 80 m

Correct Answer

Option D

Solution

Assume

a

be the retardation for both the vehicle then In case of automobile,

u_1^2 - 2a{s_1} = 0

\Rightarrow u_1^2 = 2a{s_1}

And in case for car,

u_2^2 = 2a{s_2}

$\therefore$

{\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}

\Rightarrow {\left( {{{120} \over {60}}} \right)^2} = {{{s_2}} \over {20}}

$\Rightarrow$ s2 = 80 m

Q85

The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is

A 2bv3

B -2abv2

C 2av2

D -2av3

Correct Answer

Option D

Solution

The relationship between time $t$ and distance $x$ is defined as $t = ax^2 + bx$ , where $a$ and $b$ are constants.

To find the acceleration, follow these steps: First, differentiate the equation with respect to time $t$ : $\dfrac{d}{dt}(t) = a \dfrac{d}{dt}(x^2) + b \dfrac{dx}{dt}$ This simplifies to: $1 = a \cdot 2x \dfrac{dx}{dt} + b \dfrac{dx}{dt}$ Given $\dfrac{dx}{dt} = v$ (where $v$ is the velocity), we can write: $1 = 2axv + bv = v(2ax + b)$ This simplifies to: $2ax + b = \dfrac{1}{v}$ Next, differentiate the expression $2ax + b = \dfrac{1}{v}$ again with respect to $t$ : $2a \dfrac{dx}{dt} + 0 = -\dfrac{1}{v^2} \dfrac{dv}{dt}$ Using $\dfrac{dx}{dt} = v$ , we get: $2av = -\dfrac{1}{v^2} \dfrac{dv}{dt}$ Solving for $\dfrac{dv}{dt}$ : $\dfrac{dv}{dt} = -2a v^3$ Since acceleration $f = \dfrac{dv}{dt}$ , we have: $f = -2av^3$

Q86

Two cars are travelling towards each other at speed of

20 \mathrm{~m} \mathrm{~s}^{-1}

each. When the cars are

300 \mathrm{~m}

apart, both the drivers apply brakes and the cars retard at the rate of

2 \mathrm{~m} \mathrm{~s}^{-2}

. The distance between them when they come to rest is :

A 25 m

B 100 m

C 50 m

D 200 m

Correct Answer

Option B

Solution

Let's analyze the given information before determining the distance between the two cars when they come to rest.

Each car is traveling towards the other at a speed of

20 \, \text{m s}^{-1}

and they both start braking when they are

300 \, \text{m}

apart. The deceleration (negative acceleration) of each car is given as

2 \, \text{m s}^{-2}.

To find the distance each car travels before coming to rest, we can use the kinematic equation that relates initial velocity (

v_i

), final velocity (

v_f

), acceleration (

a

), and distance (

d

), which is:

v_f^2 = v_i^2 + 2ad

Since the final velocity

v_f = 0

(they come to rest), we can rearrange the equation to solve for

d

(the distance each car travels before stopping):

0 = v_i^2 + 2ad \Rightarrow d = -\frac{v_i^2}{2a}

Plugging in the values for each car (noting that acceleration

a

is negative because it is deceleration, so

a = -2 \, \text{m s}^{-2}

):

d = -\frac{(20)^2}{2(-2)} = -\frac{400}{-4} = 100 \, \text{m}

Each car travels

100 \, \text{m}

before coming to rest. Since they both start

300 \, \text{m}

apart and each travels

100 \, \text{m}

towards the other, the total distance covered by both cars before stopping is

2 \times 100 \, \text{m} = 200 \, \text{m}

.

To find the distance between them when they come to rest, we subtract the total distance covered by both cars from the original distance between them:

300 \, \text{m} - 200 \, \text{m} = 100 \, \text{m}

So, the distance between the cars when they come to rest is

100 \, \text{m}

. Therefore, the correct option is: Option B 100 m

Q87

The co-ordinates of a moving particle at any time 't' are given by x =

\alpha

t3 and y = βt3. The speed to the particle at time 't' is given by

A

3t\sqrt {{\alpha ^2} + {\beta ^2}}

B

3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}}

C

{t^2}\sqrt {{\alpha ^2} + {\beta ^2}}

D

\sqrt {{\alpha ^2} + {\beta ^2}}

Correct Answer

Option B

Solution

Given that

x = \alpha {t^3}\,\,\,\,

and

\,\,\,\,y = \beta {t^3}

$\therefore$

{v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,

and

\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}

$\therefore$

v = \sqrt {v_x^2 + v_y^2}

= \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}}

= 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}}

Q88

Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

A 4.18 m

B 2.94 m

C 2.45 m

D 7.35 m

Correct Answer

Option D

Solution

H =

{1 \over 2}

gt2

{{9.8 \times 2} \over {9.8}}

= t2 t =

\sqrt 2

sec

\Delta

t : time interval between drops h =

{1 \over 2}

g(

\sqrt 2

$-$ 2

\Delta

t)2

\Delta

t =

{1 \over {\sqrt 2 }}

h =

{1 \over 2}

g

{\left( {\sqrt 2 - {1 \over {\sqrt 2 }}} \right)^2} = {1 \over 2} \times 9.8 \times {1 \over 2} = {{9.8} \over 4} = 2.45

m H $-$ h = 9.8 $-$ 2.45 = 7.35 m

Q89

The distance travelled by an object in time

t

is given by

s=(2.5) t^{2}

. The instantaneous speed of the object at

\mathrm{t}=5 \mathrm{~s}

will be:

A

5 \mathrm{~ms}^{-1}

B

12.5 \mathrm{~ms}^{-1}

C

62.5 \mathrm{~ms}^{-1}

D

25 \mathrm{~ms}^{-1}

Correct Answer

Option D

Solution

The distance traveled by an object in time

t

is given by the equation

s = (2.5)t^2

.

To find the instantaneous speed at a specific time, we need to find the first derivative of the distance function with respect to time, which gives us the velocity function:

v(t) = \frac{ds}{dt}

Differentiating the given equation with respect to

t

:

v(t) = \frac{d}{dt} (2.5)t^2 = 2(2.5)t = 5t

Now, we can find the instantaneous speed at

t = 5 \mathrm{~s}

by plugging the value into the velocity function:

v(5) = 5(5) = 25 \mathrm{~m/s}

The instantaneous speed of the object at

t = 5 \mathrm{~s}

is

25 \mathrm{~m/s}

.

Q90

Speeds of two identical cars are

u

and

4

u

at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is :

A

1:1

B

1:4

C

1:8

D

1:16

Correct Answer

Option D

Solution

Given the initial speeds of two identical cars as $u$ and $4u$ , and considering that both cars eventually stop (final speed $v = 0$ ), we note that both cars decelerate with the same acceleration $-a$ .

Using the kinematic equation: $v^2 = u^2 - 2as$ Since $v = 0$ , $0 = u^2 - 2as$ Hence, $u^2 = 2as$ For the first car with speed $u$ , $u^2 = 2a s_1 \quad \text{...(i)}$ For the second car with speed $4u$ , $(4u)^2 = 2a s_2 \quad \text{...(ii)}$ Dividing equation (i) by equation (ii), $\dfrac{u^2}{(4u)^2} = \dfrac{2a s_1}{2a s_2}$ $\dfrac{u^2}{16u^2} = \dfrac{s_1}{s_2}$ $\dfrac{1}{16} = \dfrac{s_1}{s_2}$ Thus, the ratio of the stopping distances of the two cars is $\dfrac{1}{16}$ .