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MCQEasyJEE 2023Velocity & Acceleration

JEE Physics 2023 Question with Solution

The position vector of a particle related to time tt is given by

r⃗=(10t i^+15t2 j^+7t k^) m.\vec r = (10t\,\hat{i}+15t^2\,\hat{j}+7t\,\hat{k})\ m.

The direction of net force experienced by the particle is

  • A

    Positive xx-axis

  • B

    Positive yy-axis

  • C

    Positive zz-axis

  • D

    In x–yx–y plane

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The position vector is

r⃗=(10t i^+15t2 j^+7t k^) m\vec r = (10t\,\hat{i}+15t^2\,\hat{j}+7t\,\hat{k})\, m

Find: The direction of the net force on the particle.

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

v⃗=dr⃗dt=(10 i^+30t j^+7 k^)\vec v=\frac{d\vec r}{dt} =\left(10\,\hat{i}+30t\,\hat{j}+7\,\hat{k}\right)

Acceleration is the derivative of velocity with respect to time:

a⃗=dv⃗dt=(0 i^+30 j^+0 k^)\vec a=\frac{d\vec v}{dt} =\left(0\,\hat{i}+30\,\hat{j}+0\,\hat{k}\right)

According to Newton’s second law,

F⃗=ma⃗\vec F=m\vec a

So, the net force acts in the same direction as the acceleration. Since acceleration has only a positive yy-component, the net force is directed along the positive yy-axis.

Therefore, the correct option is B.

Common mistakes

  • Finding only the velocity and using its direction as the direction of force is incorrect because net force depends on acceleration, not velocity. Differentiate once more to get a⃗\vec a first.

  • Assuming the force lies in the x–yx–y plane because the position vector has both xx and yy components is wrong. The direction of force is determined by the components of acceleration, not position.

  • Missing that the xx and zz components are linear in tt leads to a wrong nonzero acceleration in those directions. The second derivative of 10t10t and 7t7t is zero.

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