JEE PYQ: Motion in a Straight Line - Question ID e5a7eb16c49a (JEE Main 2023)

Core Formula & Concept:

In kinematics, the instantaneous speed of an object is defined as the magnitude of its instantaneous velocity. When the position (or distance travelled) \( s \) of an object is given as a function of time \( t \), the instantaneous velocity \( v(t) \) is obtained by differentiating \( s(t) \) with respect to \( t \):

$v(t) = \frac{ds}{dt}$

Here, \( s(t) = (2.5) t^{2} \) is the distance travelled as a function of time. The instantaneous speed at any time \( t \) is simply the absolute value of \( v(t) \), since speed is a scalar quantity and always non-negative.

Step-by-Step Derivation:

Step 1: Write down the given distance-time relation.

$s(t) = 2.5 \, t^{2}$

Step 2: Differentiate \( s(t) \) with respect to \( t \) to find the velocity function \( v(t) \).

Using the power rule of differentiation: $\frac{d}{dt} \left( t^{n} \right) = n t^{n-1}$ Apply this to \( s(t) \): $v(t) = \frac{ds}{dt} = \frac{d}{dt} \left( 2.5 \, t^{2} \right) = 2.5 \cdot 2 \, t^{2-1} = 5 \, t$

Step 3: Evaluate the velocity at \( t = 5 \, \text{s} \).

Substitute \( t = 5 \) into \( v(t) \): $v(5) = 5 \cdot 5 = 25 \, \text{ms}^{-1}$

Step 4: Interpret the result.

Since speed is the magnitude of velocity, and \( v(5) = 25 \, \text{ms}^{-1} \) is already positive, the instantaneous speed at \( t = 5 \, \text{s} \) is: $\boxed{25 \, \text{ms}^{-1}}$

Step 5: Match with the given options.

The correct option is D: \( 25 \, \text{ms}^{-1} \). Common Traps & Exam Tip:

Trap 1: Confusing distance with speed.
Some students mistakenly think that substituting \( t = 5 \) directly into \( s(t) \) gives speed. This yields \( s(5) = 2.5 \cdot 25 = 62.5 \, \text{m} \), which is distance, not speed. Speed is the rate of change of distance, not the distance itself.

Trap 2: Forgetting to differentiate.
A common oversight is to skip differentiation and assume speed equals the coefficient of \( t^2 \). This leads to incorrect values like 2.5 or 5.

Trap 3: Misapplying units.
Ensure that the final answer is in \( \text{ms}^{-1} \), not \( \text{m} \) or \( \text{s}^{-2} \). Differentiating \( s(t) \) in meters gives velocity in \( \text{ms}^{-1} \).

Exam Tip:
Always remember: Speed at an instant is the derivative of distance with respect to time. If the distance function is quadratic in \( t \), the speed function will be linear. This pattern is common in kinematics problems.