Consider the motion of a object able to move along a line under constant acceleration a. At time t = 0, its initial position and velocity are x0 and v0 respectively. Show that at a later time t, v = v0 + at, x = x0 + v0t + 12at2. Eliminate the time between these two relations to show that a(x − x0) = 12(v2 − v20). Use this to find the minimum stopping distance for a car travelling at 100kmh-1 if its maximum deceleration is 10 ms-2
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