Two cars are traveling along a straight line in the same direction,the lead car

Question

Two cars are traveling along a straight line in the same direction,the lead car at 24.0m/s and the other car at 29.8m/s. At the momentthe cars are 40.9m apart, the lead driver applies the brakes,causing her car to have an acceleration of -1.95m/s2.How long does it take for the lead car to stop? What is thedistance it travels during this time? Assuming that the chasing carbrakes at the same time as the lead car, what must be the chasingcar's minimum negative acceleration so as not to hit the leadcar?How long does it take for the chasing car to stop?

Explanation / Answer

The lead car is at VL = 24 m/s, and applies an acceleration aL= -1.95 m/s2 Using the formula Vf = VL + aLt 0 = 24 -1.95t t = 12.30769seg This is the time in which the lead car will stop. The zero inthe equation is the final velocity. The distance, we find with the formula d = VLt + (1/2)at2 VL = 24 m/s and t = 12.30769 seg d = 147.692m This is the distance that the lead car went from the point inwhich applied the break to the stop. The chasing car has to stop completly in a distance of 40.9 m+ 147.692 m = 188.59 m, and it has an initial velocity of vi2 = 29.8 m/seg. It has to reach v = 0 m/seg within thisdistance. So using V2 = vi22+2aDapart 0 = 29.8(29.8) + 2a(188.592) a= -2.3543m/seg2 From V = Vi2 + at 0 = 29.8(29.8) + 2a(188.592) a= -2.3543m/seg2 From V = Vi2 + at 0 = 29.8 -2.3543t we obtain t = 12.6576seg This is the time neededfor the chasing car to stop. This is the time neededfor the chasing car to stop.

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