# A driver of a car travelling at a velocity v suddenly sees a broad wall in front of him at a distance r , Is it better to brake or to turn sharply ?

Q: A driver of a car travelling at a velocity v suddenly sees a broad wall in front of him at a distance r , Is it better to brake or to turn sharply ?

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Sol: Let m = mass of the car . When driver applies the brake , distance covered by car before stops = x

The Retarding force $F = m a = \frac{m v^2}{2 x}$

or , $x = \frac{m v^2}{2 F}$

There will be no collision if x ≤ r

$\frac{m v^2}{2 F} \le r$

$F \ge \frac{m v^2}{2 r}$ ….(i)

If driver takes a sharp turn of radius x , then centripetal force on car is

$F’ = \frac{m v^2}{x}$

$x = \frac{m v^2}{F’}$

For no collision , x ≤ r

$\frac{m v^2}{F’} \le r$

$F’ \ge \frac{m v^2}{r}$ …(ii)

From (i) & (ii) F = F’/2

Hence to avoid collision braking force required is half the centripetal force . That means braking is better.