What average force is required to stop a 1600 kg car in 9.0 s if the car is traveling at 80 km/h?

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Dwight Share
Jan 23, 2018


First find the acceleration, then apply Newton's second law to show that the force is -3950 N.


First, we must calculate the acceleration of the car, but before we begin, we should get the initial speed into SI base units:

#80 (km)/h xx 1000 m/(km) -: 3600 s/h = 22.2 m/s#

(You may find it easier to just remember that to change km/h to m/s you divide by 3.6)

Now, the acceleration

#a = (Deltav)/(Deltat) = (v_f - v_i)/(Deltat)=(0-22.2)/9.0 = -2.47 m/s^2#

With this, we are able to apply Newton's second law:

#F_"net" = mxxa = (1600 "kg")(-2.47 m/s^2) = -3950 N#

The negative sign means that the force acts in a direction opposite the that which the car is travelling.

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