A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you run toward it with acceleration of 0.96 m/s^2. How far would you have to run before you catch up with the rear of the bus, and how fast must be running then?

Answer 1

5.00 t + 12.0 = 1/2 * 0.96 * t^2

0.48 t^2 - 5.00 t - 12 = 0

solve the quadratic for t (the positive value)

a * t is the speed at catch-up

1/2 a t^2 is the distance

Answer 2

Great!

i got 0,-12

what is (a)?

for the speed to catch-up is a = 5, or 12?

for the distance (a) should be 0.96 right?

Answer 3

never mind, got it. Thanks btw :)