Motion in a Straight Line — Page 6 | NEET Physics

Q51

Motion of a particle is given by equation s = (3t 3 + 7t 2 + 14t + 8) m. The value of acceleration of the particle at t = 1 sec is

A 10 m/s 2

B 32 m/s 2

C 23 m/s 2

D 16 m/s 2

Correct Answer

Option B

Solution

Given, s = (3t 3 + 7t 2 + 14t + 8)

{{ds} \over {dt}}

= 9t 2 + 14t + 14

{{{d^2}s} \over {d{t^2}}}

= 18t + 14

a

= 18t + 14 At t = 1, a = 18 $\times$ 1 + 14 = 32 m/s 2

Q52

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

Q53

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

Q54

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}}

Q55

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

Q56

Given below are two statements Statement I : Area under velocity- time graph gives the distance travelled by the body in a given time. Statement II : Area under acceleration- time graph is equal to the change in velocity- in the given time. In the light of given statements, choose the correct answer from the options given below.

A Both Statement I and Statement II are False.

B Both Statement I and Statement II are true.

C Statement I is incorrect but Statement II is true.

D Statement I is correct but Statement II is false.

Correct Answer

Option C

Solution

Area under velocity time graph gives displacement of body in given time.

Area under acceleration time graph gives change in velocity in the given time.

So Statement I false but Statement II True

Q57

A particle is moving in a straight line. The variation of position '

x

' as a function of time '

t

' is given as

x=\left(t^3-6 t^2+20 t+15\right) m

. The velocity of the body when its acceleration becomes zero is :

A 6 m/s

B 10 m/s

C 8 m/s

D 4 m/s

Correct Answer

Option C

Solution

The position equation is given by:

x = t^3 - 6t^2 + 20t + 15

First, compute the velocity

v

by differentiating the position function with respect to time:

\frac{d x}{d t} = v = 3t^2 - 12t + 20

Next, compute the acceleration

a

by differentiating the velocity function with respect to time:

\frac{d v}{d t} = a = 6t - 12

We need to find the time

t

when the acceleration is zero:

6t - 12 = 0

t = 2 \, \mathrm{sec}

Now, find the velocity at

t = 2 \, \mathrm{sec}

:

v = 3(2)^2 - 12(2) + 20

v = 8 \, \mathrm{m/s}

Q58

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}}}

Q59

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

.

Q60

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: $x = v_0 + \dfrac{g}{2} + \dfrac{F}{3}$