QJEEM21Ph0ccb: JEE Main 2021 Physics - Class 11 JEEM21Ph0ccb Solution: The instantaneous velocity of a particle moving in a straight…

Question

Verified Solution • 2025–26 Session moderate

The instantaneous velocity of a particle moving in a straight line is given as

V = \alpha t + \beta {t^2}

, where

\alpha

and

\beta

are constants. The distance travelled by the particle between 1s and 2s is :

Accepted Answer

V = \alpha t + \beta {t^2}

{{ds} \over {dt}} = \alpha t + \beta {t^2}

\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} }

{S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2

As particle is not changing direction

So distance = displacement

Distance =

\left[ {{{\alpha [4 - 1]} \over 2} + {{\beta [8 - 1]} \over 3}} \right]

= {{3\alpha } \over 2} + {{7\beta } \over 3}

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