The distance travelled by an object in time t is given by s = (2.… — Solution | JEE Physics

MCQEasyJEE 2023Velocity & Acceleration

The distance travelled by an object in time $t$ is given by $s = (2.5)t^2$ . The instantaneous speed of the object at $t = 5 \, \text{sec}$ will be:

Answer

Correct answer:A

Standard Method

Given: The distance travelled is $s = 2.5t^2$ .

Find: The instantaneous speed at $t = 5 \, \text{sec}$ .

Instantaneous speed is the rate of change of distance with respect to time.

v = \frac{ds}{dt} = 5t

Now substitute $t = 5$ :

v = 5 \times 5 = 25 \, \text{m/s}

Therefore, the instantaneous speed at $t = 5 \, \text{sec}$ is $25 \, \text{m/s}$ . The correct option is A.

Derivative Idea

Given: $s = 2.5t^2$

Find: Instantaneous speed at $t = 5$ .

Use the idea that instantaneous speed is the first derivative of the position function with respect to time.

\frac{d}{dt}(2.5t^2) = 5t

Then at $t = 5$ :

5t = 5(5) = 25 \, \text{m/s}

Therefore, the answer is $25 \, \text{m/s}$ .

Using average speed instead of instantaneous speed. This is wrong because the question asks for speed at a specific instant, so differentiate $s(t)$ first and then substitute the time value.
Substituting $t = 5$ directly into $s = 2.5t^2$ and treating the result as speed. This gives distance, not speed. First find $v = \frac{ds}{dt}$ .
Differentiating $2.5t^2$ incorrectly as $2.5t$ . The power rule gives $\frac{d}{dt}(t^2) = 2t$ , so the correct derivative is $5t$ .

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