# What average force is required to stop a 1500 kg car in 9.0 s if the car is traveling at 95 km/h ?

Feb 28, 2018

I got $4400 N$

#### Explanation:

We can use the Impulse-Change in Momentum Theorem:

${F}_{a v} \Delta t = \Delta p = m {v}_{f} - m {v}_{i}$

so we get:

${F}_{a v} = \frac{m {v}_{f} - m {v}_{i}}{\Delta t} = \frac{1500 \cdot 0 - 1500 \cdot 26.4}{9} = - 4400 N$ opposite to the motion direction.

where I changed $\frac{k m}{h}$ into $\frac{m}{s}$.

Feb 28, 2018

${F}_{a v e} = 4398.148148 k g \left(\frac{m}{{s}^{2}}\right)$ or $N$

#### Explanation:

From Newton's second law of motion,
$F = m a$

where F is the force, a is the acceleration and m is the mass
acceleration = $\frac{{v}_{f} - {v}_{i}}{t}$

where ${v}_{f}$ is the final velocity and ${v}_{i}$ is the initial velocity
and $t$ is the time in seconds.

So,

$F = \left(\frac{{v}_{f} - {v}_{i}}{t}\right) \cdot m$
$F = \left(\frac{0 - \left(26.38888889\right) \frac{k m}{s}}{9 s}\right) \cdot 1500 k g = - 4398.148148 N$

The minus sign indicates the direction of the force is to the opposite side

*I have converted its speed from $95 \frac{k m}{h}$ to $26.38888889 \frac{m}{s}$.