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

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Answer 1 · 2018-02-28T12:17:38

I got #4400N#

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 · 2018-02-28T12:38:58

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

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#.