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

2 Answers
Feb 28, 2018

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.

Feb 28, 2018

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.