Motion in a Straight Line — Page 6 | JEE Physics

Q51

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

Q52

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

Q53

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: $f = -2av^3$

Q54

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

Q55

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}}

Q56

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

Q57

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}

.

Q58

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 $\dfrac{1}{16}$ .

Q59

The position of a particle related to time is given by

x=\left(5 t^{2}-4 t+5\right) \mathrm{m}

. The magnitude of velocity of the particle at

t=2 s

will be :

A

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

B

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

C

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

D

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

Correct Answer

Option B

Solution

The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$ .

To find the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ , we first need to find the velocity of the particle as a function of time.

The velocity $v$ is the time derivative of the position $x$ :

v = \frac{dx}{dt}

Taking the derivative of $x$ with respect to $t$ , we get:

v = \frac{dx}{dt} = 10t - 4\,\mathrm{m/s}

Now we can find the velocity of the particle at $t=2\,\mathrm{s}$ by plugging in $t=2$ :

v(2\,\mathrm{s}) = 10(2) - 4\,\mathrm{m/s} = 16\,\mathrm{m/s}

Therefore, the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ is:

\boxed{|v(2\,\mathrm{s})| = 16\,\mathrm{m/s}}

Q60

A particle moves from the point

\left( {2.0\widehat i + 4.0\widehat j} \right)

m, at t = 0, with an initial velocity

\left( {5.0\widehat i + 4.0\widehat j} \right)

ms

-

1. It is acted upon by a constant force which produces a constant acceleration

\left( {4.0\widehat i + 4.0\widehat j} \right)

ms

-

2. What is the distance of the particle from the origin at time 2 s?

A 15 m

B

20\sqrt 2

m

C

10\sqrt 2

m

D 5 m

Correct Answer

Option B

Solution

\overrightarrow S = \left( {5\widehat i + 4} \right)2 + {1 \over 2}\left( {4\widehat i + 4\widehat j} \right)4

= 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j

\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j

\overrightarrow {{r_f}} = 20\widehat i + 20\widehat j

\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2