Question

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

Answer 1

I got `4400N`

Explanation:

We can use the Impulse-Change in Momentum Theorem:

`F_(av)Deltat=Deltap=mv_f-mv_i`

so we get:

`F_(av)=(mv_f-mv_i)/(Deltat)=(1500*0-1500*26.4)/9=-4400N` opposite to the motion direction.

where I changed `(km)/h` into `m/s`.

Answer 2

`F_(ave)= 4398.148148kg((m)/(s^2))` or `N`

Explanation:

From Newton's second law of motion,
`F=ma`

where F is the force, a is the acceleration and m is the mass
acceleration = `(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=((v_f-v_i)/t)*m`
`F=((0-(26.38888889)(km)/s)/(9s))*1500 kg = -4398.148148N`

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

*I have converted its speed from `95(km)/h` to `26.38888889(m)/s`.