Bike and Car both accelerate from rest ( intermediate problem)

[Color_of_Cyan]

Bike and Car both accelerate from rest ("intermediate" problem)

Homework Statement

Basically, both a bike and car start from rest. The bike will not go any faster than (final velocity) 20 mi / hr. The car will not go faster than (final velocity) 50 mi / hr. However, at first, the bike is ahead of the car because it has a constant acceleration of 13 mi / (hr s), which is higher than that of the car's constant acceleration of 9 mi / (hr s).


Part a: What is the time it takes for the car to catch up to (and have the same position of) the bicycle.

Part b: What is the MAXIMUM DISTANCE of the bike being ahead of the car before the car has the same velocity of the bike and starts to catch up?

Homework Equations

Vf = Vi + at

Vf² = Vi² + 2a(Xf - Xi)

Xf = Xi + Vi t + 1/2 at²

etc

The Attempt at a Solution

initial velocity of car and bike = 0

acceleration of bike = 13 mi / hr s

FINAL velocity of bike = 20 mi / hr s

FINAL velocity of car = 50 mi / hr s

Is it possible to solve this problem without the use of calculus? If I do, how would I apply it to here. Anyways, for part a if I substitute Xf = Xi + Vi t + 1/2 at² (plugging whatever is the acceleration of the bike) equal to xf = (whatever is the acceleration of the car), then I would just end up with t as a variable and then t would just cancel out, which I do not want.

However for the car, using vf = vi + at, and then 20 mi / hr = 0 + [(20 mi / (hr s) ]t it will take the CAR 2.23 seconds to reach the SAME VELOCITY as the BIKE, but this is not what I want to solve for. I know I would probably have to improvise somewhere, but where?

 

[gneill]

I suggest that you first calculate the times and distances for certain "events" that take place, namely the times and distances from the start when the bike and car switch over from accelerated motion to constant velocity motion. What are the positions of car and bike at each of these events?