A 2989 kg car moves down a level highway under the acceleration of two forces. One is 1182 N forward force expected on the drive while by the road; the other is 901 N resistive force. Answer in units of m/s

Please note that, since the two forces are opposite each other, then the magnitude a of the acceleration of your car, is:

a=(1182-901)/2989=0.094 m/sec^2

Since your car has traveled a distance d, I can apply this formula:

d= (1/2) a* t^2,

where t is the time needed to our car to travel that distance d.

From the last formula, I get t as below:

t= sqrt[(2d/a)]

then I substitute into the formula which gives the speed v reachd by our car, namely:

v = a*t =a* sqrt[(2d)/a], so

v=sqrt (2 d a)

using your numerical dat, we get:

v= sqrt(2*0.094*25)=2.17 m/sec

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