×

# What average force is required to stop a 1600 kg car in 9.0 s if the car is traveling at 80 km/h?

Dec 21, 2016

First find the acceleration, then apply Newton's second law to show that the force is -3950 N.

#### Explanation:

First, we must calculate the acceleration of the car, but before we begin, we should get the initial speed into SI base units:

$80 \frac{k m}{h} \times 1000 \frac{m}{k m} \div 3600 \frac{s}{h} = 22.2 \frac{m}{s}$

(You may find it easier to just remember that to change km/h to m/s you divide by 3.6)

Now, the acceleration

$a = \frac{\Delta v}{\Delta t} = \frac{{v}_{f} - {v}_{i}}{\Delta t} = \frac{0 - 22.2}{9.0} = - 2.47 \frac{m}{s} ^ 2$

With this, we are able to apply Newton's second law:

F_"net" = mxxa = (1600 "kg")(-2.47 m/s^2) = -3950 N

The negative sign means that the force acts in a direction opposite the that which the car is travelling.