a physics student ‘borrows’ a sports car for a joy ride and discovers that it can accelerate at a rate of 4.90 m/s2. He decides to test the car by challenging Mr. Horn and his motorcycle. Both start from rest, but the student is so confident in his new ride that he gives Mr. Horn a 1.00 s head start. If Mr. Horn moves with a constant acceleration of 3.50 m/s2 and the student maintains his acceleration of 4.90 m/s2, find:

Question

(a) the time it takes the student to overcome Mr. Horn.
(b) the distance he travels before he catches up with Mr. Horn.
(c) the speed of both vehicles at the instant the student overtakes Mr. Horn.

Answer 1

recall that
v = at
s = 1/2 at^2
the distances traveled at time t are
Horn: 3.50/2 t^2
student: 4.9/2 (t-1)^2
so now you can answer the questions

Answer 2

Mr. Horn travels 1.75 m during his 1 sec head start
... his velocity is 3.50 m/s when the student starts

(a) t is the time the student is moving ... Mr. Horn's time is 1 sec greater
... 4.90 t^2 = 3.50 t^2 + 3.50 t + 1.75 ... 1.40 t^2 - 3.50 t - 1.75 = 0
... t = 2.93 s

(b) d = 1/2 a t^2 = 1/2 * 4.90 * 2.93^2

(c) v = a t
... student ... v = 4.90 m/s^2 * 2.93 s
... Mr. Horn ... v = 3.50 m/s^2 * 3.93 s

Answer 3

oops ... didn't adjust the coefficients for Mr. Horn's travel

(a) 4.90 t^2 = 3.50 t^2 + 7.00 t + 3.50 ... 1.40 t^2 - 7.00 t - 3.50 = 0
... t = 5.46 s

plug the new time into (b) and (c)

sorry about that