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MCQEasyJEE 2023Velocity & Acceleration

JEE Physics 2023 Question with Solution

The position of a particle related to time is given by x=(5t2−4t+5) m.x=(5t^2-4t+5)\ \text{m}. The magnitude of velocity of the particle at t=2 st=2\,s will be

  • A

    14 m s−114\ \text{m s}^{-1}

  • B

    16 m s−116\ \text{m s}^{-1}

  • C

    10 m s−110\ \text{m s}^{-1}

  • D

    6 m s−16\ \text{m s}^{-1}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The position function is x(t)=5t2−4t+5x(t)=5t^2-4t+5.

Find: The magnitude of velocity at t=2 st=2\,s.

Velocity is the time derivative of position.

v(t)=dxdtv(t)=\frac{dx}{dt}

So,

v(t)=ddt(5t2−4t+5)v(t)=\frac{d}{dt}(5t^2-4t+5) v(t)=10t−4v(t)=10t-4

Now substitute t=2 st=2\,s:

v(2)=10(2)−4=20−4=16 m s−1v(2)=10(2)-4=20-4=16\ \text{m s}^{-1}

Since the value is already positive, its magnitude remains the same.

∣v∣=16 m s−1|v|=16\ \text{m s}^{-1}

Therefore, the correct option is B.

Common mistakes

  • Differentiating the position function incorrectly. Velocity is the derivative of position with respect to time, so ddt(5t2−4t+5)=10t−4\frac{d}{dt}(5t^2-4t+5)=10t-4, not 5t2−4t+55t^2-4t+5 itself.

  • Substituting t=2t=2 before differentiating. First find v(t)v(t) from x(t)x(t), then put t=2t=2 to get the velocity at that instant.

  • Ignoring that the question asks for magnitude of velocity. After finding velocity, take its modulus if needed; here the velocity is positive, so the magnitude is unchanged.

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