Physics - acceleration/velocity question?

Question

"A man runs at speed v to catch the bus. At t=0, he is still a distance of 10m from the door when the bus starts to accelerate at a constant rate of 0.38 m/s^2. What is the minimum speed the man can run and still board the bus."

I know this involves a quadratic because I assume you have to use the equation [(xf) = (xi) + (vi)(t) + (0.5)(a)(t^2)], and that [b^2 - 4ac] (from under the root in solving a quadratic using the Q.F.) must equal zero in order to ensure that the man's speed is the minimum it can be so that he and the bus are in the same place at the same time... I'm just having difficulty setting up the problem properly.

Thanks in advance!

Answer 1

Sorry I do not understand your equation: [(xf) = (xi) + (vi)(t) + (0.5)(a)(t^2)],
Assume that the man can run at a constant speed v and still board the bus at time t. Also assume that the bus accelerates away from the man instead of towards the man. We have:
vt = 10 + 0.5a t^2 = 10 + 0.19t^2
Now we want to find t which minimize possitive v.
Have you learned calculus? It says that the first derivative of v w.r.t. t must be zero in order for v to be minimized.
dv/dt = -10/t^2 + 0.19 = 0
or 0.19t^2 =10
with this (positive) t, we have:
v = (10 + 0.19t^2)/t = 20/t = 2.76 (m/s)

Answer 2

a) commencing the clock whilst highway Runner passes Wile E., how a lot time elapses earlier Wile E. catches as much as highway Runner? Dr = distance of the line Runner after time 0 Dw = Distance of Wile E Coyote after time 0 Dr = 36t Dw = 4.75t + 0.5(3.5)t^2 Wile E will capture highway Runner whilst Dr = Dw, so: 36t = 4.75t + 0.5(3.5)t^2 remedy for t. (hint, divide the two factors by t). b) How a good distance have they run whilst Wile E. catches up? merely use that fee of t with the Dr equation. c) how briskly is Wile E. working whilst he reaches his quarry the line Runner? V = Vo + At V = 4.seventy 5 + 3.5t Plug in t. d) If on the on the spot Wile E. catches as much as highway Runner he attempts to decelerate as a fashion to no longer bypass over the cliff in user-friendly terms one hundred twenty-meters added forward, then what minimum acceleration is needed for Wile E. to offer up earlier plunging down the cliff? right here you utilize a similar equations: one hundred twenty = Vo + 0.5At^2 Use the V you calculated for c), 0 = Vo + At remedy those 2 equation to discover a.