A 1000 kg car moving at 20 m/s brakes steadily and stops in 40 m…

Step 1. (a) $0 = (20\ \text{m/s})^2 - 2a \times 40\ \text{m}$, so $a = \dfrac{400\ \text{m}^2/\text{s}^2}{80\ \text{m}} = 5\ \text{m/s}^2$ (a deceleration).

Step 2. (b) $F = 1000\ \text{kg} \times 5\ \text{m/s}^2 = 5000\ \text{N}$.

Step 3. (c) $t = \dfrac{20\ \text{m/s}}{5\ \text{m/s}^2} = 4\ \text{s}$.