Question

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

The motion is a body thrown up and returning under gravity; check each graph's slope and shape against \(x=ut-\frac12gt^2\).

Answer 1

The Correct Option is C

Let the body be projected up with speed u: \(x=ut-\frac12gt^2\), \(v=u-gt\); the total distance s is the path length, which is non-negative and never decreases.
Distance-time slope \(=|v|\), which equals u at \(t=0\) (non-zero) and is zero only for an instant at the top, after which it grows again.
Graph (3) shows zero slope at \(t=0\) and a permanently flat end (speed zero for ever), which contradicts \(|v|=|u-gt|\). The other three graphs match v-t (linear), v-x (\(v^2=u^2-2gx\)) and x-t (inverted parabola).
Answer: (3)