Physics · 2021
JEE Main · 25 July 2021, Shift 2 · Q3
The instantaneous velocity of a particle moving in a straight line is given as v=α t+β t^2, where α and β are constants. The distance travelled by…
The instantaneous velocity of a particle moving in a straight line is given as $\displaystyle v=\alpha t+\beta t^2$, where $\displaystyle \alpha$ and $\displaystyle \beta$ are constants. The distance travelled by the particle between $\displaystyle 1$ s and $\displaystyle 2$ s is :
Official answer
(2)
$\displaystyle \frac{3}{2} \alpha+\frac{7}{3} \beta$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.