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MCQEasyJEE 2024Velocity & Acceleration

JEE Physics 2024 Question with Solution

A particle is moving in a straight line. The variation of position xx as a function of time tt is given as x=t3−6t2+20t+15 mx = t^3 - 6t^2 + 20t + 15 \, \text{m}. The velocity of the body when its acceleration becomes zero is:

  • A

    4 m/s4 \, \text{m/s}

  • B

    8 m/s8 \, \text{m/s}

  • C

    10 m/s10 \, \text{m/s}

  • D

    6 m/s6 \, \text{m/s}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: x=t3−6t2+20t+15 mx = t^3 - 6t^2 + 20t + 15 \, \text{m}

Find: The velocity when acceleration becomes zero.

Velocity is the first derivative of position with respect to time:

v=dxdt=ddt(t3−6t2+20t+15)v = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 20t + 15)

So,

v=3t2−12t+20v = 3t^2 - 12t + 20

Acceleration is the derivative of velocity with respect to time:

a=dvdt=ddt(3t2−12t+20)a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 20)

Hence,

a=6t−12a = 6t - 12

When acceleration becomes zero,

6t−12=06t - 12 = 0

Therefore,

6t=126t = 12 t=2t = 2

Now substitute t=2t = 2 into the velocity expression:

v=3(2)2−12(2)+20v = 3(2)^2 - 12(2) + 20 v=3(4)−24+20v = 3(4) - 24 + 20 v=12−24+20v = 12 - 24 + 20 v=8 m/sv = 8 \, \text{m/s}

Therefore, the velocity of the body when its acceleration becomes zero is 8 m/s8 \, \text{m/s}. The correct option is B.

Derivative-Based Approach

Given: The position function is x=t3−6t2+20t+15x = t^3 - 6t^2 + 20t + 15.

Find: The value of velocity at the instant when a=0a = 0.

Use the kinematics relation from calculus: position gives velocity by first differentiation, and velocity gives acceleration by second differentiation.

First differentiate position:

dxdt=ddt(t3)−ddt(6t2)+ddt(20t)+ddt(15)\frac{dx}{dt} = \frac{d}{dt}(t^3) - \frac{d}{dt}(6t^2) + \frac{d}{dt}(20t) + \frac{d}{dt}(15) v=3t2−12t+20v = 3t^2 - 12t + 20

Now differentiate velocity:

a=dvdt=ddt(3t2−12t+20)a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 20) a=6t−12a = 6t - 12

Set acceleration equal to zero:

6t−12=06t - 12 = 0 6t=126t = 12 t=2 st = 2 \, \text{s}

Evaluate velocity at this time:

v(2)=3(2)2−12(2)+20v(2) = 3(2)^2 - 12(2) + 20 =12−24+20= 12 - 24 + 20 =8 m/s= 8 \, \text{m/s}

Thus, the required velocity is 8 m/s8 \, \text{m/s}, so the answer is B.

Common mistakes

  • Differentiating the position function incorrectly. For example, taking the derivative of t3t^3 or −6t2-6t^2 wrongly gives an incorrect velocity expression. Differentiate term by term carefully to get v=3t2−12t+20v = 3t^2 - 12t + 20.

  • Using the condition v=0v = 0 instead of a=0a = 0. The question asks for velocity when acceleration becomes zero, so first set a=dvdt=0a = \frac{dv}{dt} = 0 and only then compute velocity.

  • Finding t=2t = 2 correctly but not substituting it back into the velocity equation. The time when acceleration is zero is only an intermediate result; the required quantity is velocity at that time.

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