What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h?

i got the answer to be 13062.5 N. i don't think that's right...

4 answers

Impluse= change in momentum
Force*time= 1100*26.4m/s

I don't get your numbers, about one fourth of that. Did your conversion of km/hr to m/s same as mine?

Here is a neat trick. Put in the google search window...

1100kg*95km/hr* 1/(8seconds) in newtons
First convert kg/h to m/s
95 km/h = (95000 m/h)/(3600) s/h
= 26.4 m/s

Force x Time = momentum change
= 1100*26.4 = 2.90*10^4 kg m/s

Divide that by the time interval (8.0s) to get the average force in Newtons
Change velocity from km/h to m/s
95 km/h*(1000 m/km)*(1 h/3600 s)= 26.4 m/s
a=ΔV/Δt=26.4/8=3.3 m/s^2
ΣF=m*a=1100kg*3.3 m/s^2=3630 N

Tada!
wrong, it should be -3630 since the change in velocity must be negative since it is an initial velocity