Question

A 1,000 kg car moving at a speed of $20m{s}^{-1}$​ is brought to rest at a distance of 50 m, calculate the unbalanced force acting on the car and give a reason why the actual force applied by the brakes may be slightly less than that calculated?

• A. (-) 4000 N, More
• B. 3000 N, Less
• C. 2000 N, More
• D. 1000 N, More

Answer 1

The correct option is A

(-) 4000 N, More

Step 1: Here u = $20m{s}^{-1}$​; v = 0; s = 50 m; a = ?

Using $v^2-u^2=2as$, we have

$a=\frac{v^2-u^2}{2s}$ = $-4m{s}^{-2}$

Step2: F = ma = 1,000$\times$(-4) = -4,000 N

Due to force of friction, the actual force applied by brakes may be slightly less than the calculated one.