JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 14)
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 :
3$$\alpha$$ + 7$$\beta$$
$${3 \over 2}\alpha + {7 \over 3}\beta $$
$${\alpha \over 2} + {\beta \over 3}$$
$${3 \over 2}\alpha + {7 \over 2}\beta $$
Explanation
$$V = \alpha t + \beta {t^2}$$
$${{ds} \over {dt}} = \alpha t + \beta {t^2}$$
$$\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} } $$
$${S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2$$
As particle is not changing direction
So distance = displacement
Distance = $$\left[ {{{\alpha [4 - 1]} \over 2} + {{\beta [8 - 1]} \over 3}} \right]$$
$$ = {{3\alpha } \over 2} + {{7\beta } \over 3}$$
$${{ds} \over {dt}} = \alpha t + \beta {t^2}$$
$$\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} } $$
$${S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2$$
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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