distance the student runs=38.9+4.6t
distance the bus goes=1/2 a t^2
in the same time, the respective distances are equal. Set them equal, and solve for time.
A student is running at her top speed of 4.6 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.9 m from the bus, it starts to pull away, moving with a constant acceleration of 0.177 m/s^2.
A) For how much time does the student have to run at 4.6 m/s before she overtakes the bus?
B) For what distance does the student have to run at 4.6 m/s before she overtakes the bus?
C) When she reaches the bus, how fast is the bus traveling?
D) What is the minimum speed the student must have to just catch up with the bus?
E) For what time does she have to run in that case?
F) For what distance does she have to run in that case?
2 answers
Can I use the same equations to solve for the distance in part B?