Motion in a Straight Line question for JEE Main (JEE Main 2018 (8 Apr)), with solution

Answer: (C) see figure

Graphs (1), (2) and (4) describe a body thrown vertically up (constant acceleration $-g$):

- (1) $v$ decreases linearly with time and becomes negative;
- (2) $v^2 = u^2 - 2gx$: velocity vs position is a parabola, $+u$ going up and $-u$ coming back down at the same positions;
- (4) position vs time is a downward parabola.

At $t = 0$ the body already has speed $u$, so its distance–time graph must start with a non-zero slope. Graph (3) starts with zero slope (as if starting from rest) and is the incorrect one.

Solution by Sreeraj P, M.Sc Physics