help with Physics acceleration question?

Question

The rear of a bicycle passes a point O on a road with a velocity of 4 m s-1 and an acceleration of 2 m s-2.
4 seconds later the front of a car passes O with a velocity of 2 m s-1 and an acceleration of 4 m s-2.
When and how far from O does the dront of the car meet the rear of the bicycle?

The answer in the back of my book is 16.5s and 338.25m but i have no idea how to get it...

Answer 1

eyeonthescreen got the eqs right. I carried out the operation and found the quadratic has 2 solutions; t = 1.45 sec → 7.9 m and t = 16.55 sec → 340.1 m

The physical interpretation of this is that at t = 1.45 sec, the bike has just past the car, so their distances are equal; later, at t = 16.55 sec, the car catches up and the distances are equal again. Since the problem only addresses times > 4 sec, the first time doesn't count and 16.55 sec/340.1 m is the answer.

Answer 2

u = 4 mps and a = 2 m/sec^2 at time t0 = 0 and point O

U = 2 mps and A = 4 m/sec^2 at time t1 = 4 sec and point O

The question wants to know when d = D, the distance from O for the bike's tail and the car's front are equal.

d = ut + 1/2 at^2 and D = U(t - 4) + 1/2 A(t - 4)^2 from the SUVAT equations. t - 4 represents the time the car travels relative to the time t elapsed for the biker. Set d = D; so that we have:

d = ut + 1/2 at^2 = U(t - 4) + 1/2 A(t - 4)^2 = D and solve for t. Plug that time answer into either the d or D equation to solve for d or D to get the distance from O.

Answer 3

x(bike) = 4t+t^2
x(bike)(t=4) = 32m
x(car) = 2(t-4)+2(t-4)^2

Set x(bike) = x(car) and solve for t. Then plug that value of t into either x equation to get the distance. Check your answer by making sure that you ge the same answer from either x equation.