JEE Main 2021 (17 Mar, Shift 1) Physics Q1: Kinematics in 1D

Question

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is t seconds, the total distance travelled is :

Accepted Answer

Option 2: $\frac{\alpha\beta}{2(\alpha+\beta)}t^2$

Answer

$\frac{2\alpha\beta}{(\alpha+\beta)}t^2$

Answer

$\frac{\alpha\beta}{2(\alpha+\beta)}t^2$

Answer

$\frac{4\alpha\beta}{(\alpha+\beta)}t^2$

Answer

$\frac{\alpha\beta}{4(\alpha+\beta)}t^2$

JEE Main 17 March 2021, Shift 1, Physics Q1

Same idea in other shifts