Kinematics: Application Problems (7)

Question

A car moving at $20 \text{ m/s}$ applies brakes and decelerates uniformly at $4 \text{ m/s}^2$. How far does it travel before coming to rest, and how long does the stop take? Then, if a second car follows $30 \text{ m}$ behind at the same speed and reacts $0.5 \text{ s}$ later with the same deceleration, will it crash into the first car?

This is a standard kinematics application problem that mirrors what you see in JEE Main and NEET — two-vehicle problems with reaction time built in.

Solution — Step by Step

Why This Works

Kinematics with uniform acceleration is fully described by three equations: $v = u + at$, $s = ut + \tfrac{1}{2}at^2$, and $v^2 = u^2 + 2as$. The trick in real problems is identifying which equation kills which unknown fastest. Here, the third equation gives us $s$ directly without needing $t$ — that’s why we used it first.

The reaction-time piece is just constant-velocity motion stitched onto the front of a deceleration phase. We never mix equations across the two phases — we treat them as two segments and add the displacements.

Alternative Method

You can solve this graphically using a $v$-$t$ diagram. Both cars are triangles in $v$-$t$ space (linear deceleration). Car 2’s diagram has a $0.5 \text{ s}$ rectangle of width $0.5$ and height $20$ stuck on the left, then the same triangle as car 1. The total area under each curve equals the stopping distance — and the area difference tells you exactly the safety margin.

Common Mistake

Final answer: Car 1 stops in $50 \text{ m}$ over $5 \text{ s}$. Car 2 needs $60 \text{ m}$ total, has $80 \text{ m}$ available, so no collision with $20 \text{ m}$ to spare.