Given that the brakes impart to the motorcycle a constant negative acceleration a and that the haywagon is moving with speed v1in the same direction as the motorcycle, show that the motorcyclist can avoid collision only if he is traveling at a speed less than v1+ sq.rt (2 lals).
Let v2 be the speed of the bike. The resulting speed of approach is v(0) v2-v1 at t0. So, with a deceleration of a, we have v(t) v2 - v1 - at. The distance D(t) between the two vehicles is D(t) D(0) - v(0)t - at?/2 s - (v2-v1)t + at?/2. We want D(t) to be positive when the bike slows down to the same speed as the wagon***, i.e. v(t) 0. Solving for t we get v(t) v2 - v1 - at 0 → t (v1-v2)/a Plugging this in the D(t) formula we get D(t) s - (v2-v1)?/a + a(v1-v2)?/2a? 0 → (v2-v1)?/a - (v1-v2)?/2a s → (v2-v1)?/2a s → v2-v1 sqrt(2as) → v2 v1 + sqrt(2as) *** Grizzly, there's no need for the bike to stop. Just to slow down to the same speed as the wagon is enough.
Solve to see what velocity he'll have to be going to just hit the wagon as he comes to a stop. If he's going at velocity v0 and the time it takes him to come to a stop is t, then he will travel a distance x given by x v0 * t + (1/2) * a * t^2 During that time, the wagon will travel a distance d v1 * t So if the total distance the motorcyclist travels, x, is equal to s + (v1 * t), the two will end up in the same spot. x v0 * t + (1/2) * a * t^2 s + v1 * t (call this equation 1) v0 * t + (1/2) * a * t^2 - s - (v1 * t) 0 Now gather all the terms involving similar terms of t: (1/2 * a) t^2 + (v0 - v1) t - s 0 This is a quadratic equation of the standard form Ax^2 + Bx + C, where A (1/2) * a B (v0 - v1) C -s Plug and chug through the quadratic equation to solve for t ,the length of time it will take the motorcyclist to stop, in terms of a,v0, v1, and s. You should get a positive and negative value for t; only the positive one is valid for this problem. Plug the value of t back into equation1 above, only change the sign to a sign. Lots of algebra involved, but you should end up with the right answer.
