When and where does student catch bus?

az_lender

The student's position at "t" seconds after she reaches the bus stop is (4.30 m/s)*t. The bus's position at "t" seconds after the student reaches the bus stop is (1/2)(0.952 m/s^2)(t + 2s)^2. If the student catches the bus, it happens when (4.30 m/s t) = (1/2)(0.952 m/s^2)(t + 2s)^2, or 8.60s t/0.952 = t^2 + 4s t + 4s^2, or 0 = t^2 - 5.034s t + 4s^2; use the quadratic function: t = 2.517s +/- (1/2)*sqrt[(5.034s)^2 - 16s^2] = 2.517s +/- (1/2)*sqrt(9.341 s^2) = 0.989s or 4.045s. The student catches the bus 0.989 seconds after reaching the bus stop, at which time she will have run 4.25 m beyond the bus stop.

electron1

Let’s determine the distance the bus moves in two seconds. d = vi * t + ½ * a * t^2, vi = 0 d = ½ * 0.950 * 2^2 = 1.9 meters For the bus: d = 1.9 + 0.475 * t^2 For the student: d = 4.3 * t 1.9 + 0.475 * t^2 = 4.3 * t 0.475 * t^2 – 4.3 * t + 1.9 = 0 t = [4.3 ± √(-4.3^2 – 4 * 0.475 * 1.9)] ÷ 0.95 t1 = [4.3 + √14.88] ÷ 0.95 This is approximately 8.6 seconds. t2 = [4.3 – √14.88] ÷ 0.95 This is approximately 0.465 seconds. Let’s use the first time. For the bus: d = 1.9 + 0.475 * ([4.3 + √14.88] ÷ 0.95)^2 This is approximately 36.9 meters. For the student: d = 4.3 * [4.3 + √14.88] ÷ 0.95 This is approximately 36.9 meters. I believe that this is the correct answer.

Andrew Smith

At the next bus stop. The bus isn't just going to stop and let you on. I have done this too many times so I have personal experience of overtaking the bus and arriving at the next stop before it does.