Motion in a Straight Line — Page 2 | NEET Physics

Q11

Two bodies of mass 1 kg and 3 kg dropped from 16 m and 25 m. Ratio of time taken:

A 4/5

B 5/4

C 12/5

D 5/12

Correct Answer

Option A

Solution

h = (1/2)*g*t^2. t1/t2 = sqrt(16/25) = 4/5

Q12

The displacement x = ae^(-at) + be^(bt). Velocity will:

A be independent of beta

B drop to zero when alpha = beta

C go on decreasing with time

D go on increasing with time

Correct Answer

Option D

Solution

v = -alpha*a*e^(-alpha*t) + b*beta*e^(beta*t). As t increases the second term dominates, so velocity increases.

Q13

The following plots show variation of velocity

(v)

with time

(t)

of a ball thrown vertically upward, and falling back. Which of the following plots is/are correct?

A C only

B D only

C B only

D A and E only

Correct Answer

Option A

Solution

During the whole journey, acceleration due to gravity is vertically downward.

Therefore, slope of velocity vs time curve should be negative throughout the journey.

∴ Statement (C) is correct

Q14

When a ruler falls vertically, 5 different persons catch it with different reaction times. (

g=9.8 \mathrm{~m} \mathrm{~s}^{-2}

) A. Person A has reaction time of 0.20 s . B. Person

B

has reaction time of 0.22 s . C. Person C has reaction time of 0.18 s . D. Person D has reaction time of 0.19 s . E. Person E has reaction time of 0.21 s . What is the correct order of the distance travelled by the ruler for each person?

A B

>

E

>

A

>

C

>

D

B C

>

D

>

A

>

B

>

E

C B

>

E

>

A

>

D

>

C

D C

>

D

>

A

>

E

>

B

Correct Answer

Option C

Solution

→ There will be large distance for large reaction time → Descending order of reaction time

\Rightarrow t_B>t_E>t_A>t_D>t_C

→ Descending order of distance covered

\Rightarrow S_B>S_E>S_A>S_D>S_C

Q15

In some appropriate units, time

(t)

and position

(x)

relation of a moving particle is given by

t=x^2+x

. The acceleration of the particle is

A

+\dfrac{2}{(x+1)^3}

B

+\dfrac{2}{2 x+1}

C

-\dfrac{2}{(x+2)^3}

D

-\dfrac{2}{(2 x+1)^3}

Correct Answer

Option D

Solution

Given the relationship between time $t$ and position $x$ of a moving particle: $t = x^2 + x$ First, find the derivative of $t$ with respect to $x$ : $\dfrac{d t}{d x} = 2x + 1$ The velocity $v$ is the inverse of this derivative, as it's given by the derivative of position $x$ with respect to time $t$ : $v = \dfrac{d x}{d t} = \dfrac{1}{2x + 1}$ Next, calculate the derivative of $v$ with respect to $x$ : $\dfrac{d v}{d x} = \dfrac{-2}{(2x + 1)^2}$ The acceleration $a$ is then the product of velocity $v$ and the derivative of velocity with respect to $x$ : $a = v \cdot \dfrac{d v}{d x} = \dfrac{1}{2x + 1} \cdot \left(\dfrac{-2}{(2x + 1)^2}\right)$ Simplifying this expression, we find: $a = -\dfrac{2}{(2x + 1)^3}$

Q16

Two cities

X

and

Y

are connected by a regular bus service with a bus leaving in either direction every

T

min. A girl is driving scooty with a speed of

60 \mathrm{~km} / \mathrm{h}

in the direction

X

to

Y

notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period

T

of the bus service and the speed (assumed constant) of the buses.

A

10 \mathrm{~min}, 90 \mathrm{~km} / \mathrm{h}

B

15 \mathrm{~min}, 120 \mathrm{~km} / \mathrm{h}

C

9 \mathrm{~min}, 40 \mathrm{~km} / \mathrm{h}

D

25 \mathrm{~min}, 100 \mathrm{~km} / \mathrm{h}

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Let velocity of bus }=v \mathrm{~km} / \mathrm{hr} \\ & \text{ Relative velocity of bus w.r.t. scooty }=(v-60) \\ & \text{ Distance between } 2 \text{ consecutive buses }=v T \\ & (v-60) 30=v T \quad\text{.... (i)}\\ & Y \rightarrow X \\ & (v+60) 10=v T \quad\text{.... (ii)}\\ & \text{ Equating }(1) \text{ and }(2) \\ & (v-60) 30=(v+60) 10 \\ & \therefore \quad v=120 \mathrm{~km} / \mathrm{hr} \\ & T=15 \mathrm{~min} \end{aligned}

Q17

A particle is moving along

x

-axis with its position (x) varying with time

(t)

as

x=\alpha t^4+\beta t^2+\gamma t+\delta

. The ratio of its initial velocity to its initial acceleration, respectively, is:

A

2 \alpha: \delta

B

\gamma: 2 \delta

C

4 \alpha: \beta

D

\gamma: 2 \beta

Correct Answer

Option D

Solution

To find the ratio of the initial velocity to the initial acceleration of a particle moving along the

x

-axis, we need to differentiate the given position function with respect to time

t

. Let's start by writing the position function:

x = \alpha t^4 + \beta t^2 + \gamma t + \delta

To find the velocity (

v

), we differentiate

x

with respect to

t

:

v = \frac{dx}{dt} = \frac{d}{dt} (\alpha t^4 + \beta t^2 + \gamma t + \delta)

This gives us:

v = 4\alpha t^3 + 2\beta t + \gamma

The initial velocity is the velocity at

t = 0

:

v(0) = 4\alpha (0)^3 + 2\beta (0) + \gamma = \gamma

Next, to find the acceleration (

a

), we differentiate

v

with respect to

t

:

a = \frac{dv}{dt} = \frac{d}{dt} (4\alpha t^3 + 2\beta t + \gamma)

This gives us:

a = 12\alpha t^2 + 2\beta

The initial acceleration is the acceleration at

t = 0

:

a(0) = 12\alpha (0)^2 + 2\beta = 2\beta

Now, we find the ratio of the initial velocity to the initial acceleration:

\text{Ratio} = \frac{v(0)}{a(0)} = \frac{\gamma}{2\beta}

Therefore, the correct answer is: Option D:

\gamma: 2\beta

Q18

The position of a particle is given by

\vec{r}(t)=4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k}

where

\mathrm{t}

is in seconds and

\mathrm{r}

in metre. Find the magnitude and direction of velocity

v(t)

, at

t=1 \mathrm{~s}

, with respect to

\mathrm{x}

-axis

A

4 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}

B

4 \sqrt{2} \mathrm{~ms}^{-1}, 60^{\circ}

C

3 \sqrt{2} \mathrm{~ms}^{-1}, 30^{\circ}

D

3 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}

Correct Answer

Option A

Solution

\begin{aligned} & \overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\ & \text{ at } \mathrm{t}=1 \mathrm{sec} \\ & \overrightarrow{\mathrm{V}}=4 \hat{\mathrm{i}}+4(1) \hat{\mathrm{j}} \\ & |\overrightarrow{\mathrm{V}}|=\sqrt{4^2+4^2}=4 \sqrt{2} \\ & \tan \alpha=\frac{4}{4}=1 \\ & \alpha=45^{\circ}\end{aligned}

Q19

A vehicle travels half the distance with speed

v

and the remaining distance with speed

2 v

. Its average speed is :

A

\dfrac{2 v}{3}

B

\dfrac{4 v}{3}

C

\dfrac{3 v}{4}

D

\dfrac{v}{3}

Correct Answer

Option B

Solution

V_{\text{avg }}=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}=\frac{2(v)(2 v)}{v+2 v}=\frac{4 v^{2}}{3 v}=\frac{4 v}{3}

Q20

The ratio of the distances travelled by a freely falling body in the 1 st , 2 nd , 3 rd and 4 th second

A 1 : 2 : 3 : 4

B 1 : 4 : 9 : 16

C 1 : 3 : 5 : 7

D 1 : 1 : 1 : 1

Correct Answer

Option C

Solution

{S_{{n^{th}}}} = u + {1 \over 2}a(2n - 1)

{S_{{1^{st}}}} = {1 \over 2}g(2 \times 1 - 1) = {g \over 2}

{S_{{2^{nd}}}} = {1 \over 2}g(2 \times 2 - 1) = 3 \times \left( {{1 \over 2}g} \right)

{S_{{3^{rd}}}} = {1 \over 2}g(2 \times 3 - 1) = 5 \times \left( {{1 \over 2}g} \right)

{S_{{4^{th}}}} = {1 \over 2}g(2 \times 4 - 1) = 7 \times \left( {{1 \over 2}g} \right)

{S_{{1^{st}}}}:{S_{{2^{nd}}}}:{S_{{3^{rd}}}}:{S_{{4^{th}}}}

= 1:3:5:7