1. ## car and bike

a cyclist travelling at 8 m/s^-1 observes a a car just beginning to accelerate uniformly 40 m distant. if the cyclist speed remains constant he can just catch the car. find the acceleration of the car

trying to solve this using uvast equations. cyclist u=8 v=8 s=40+s a=o t=t car u=0 s=s a=a t=t there are 3 variables here so i must be missing something

2. ## Re: car and bike

Do you understand what "just catch the car" means? Assuming that the car is initially stopped, and taking t to be 0 at the moment the car starts accelerating, and taking the distance to be 0 where the bicycle is at that time, then, for the bicycle, s= 8t m. For the car, s= (1/2)(a)t^2+ 40 m. The bicyclist will "just catch the car" (but not pass it) if the two distances (from the original position of the bicyclist) are the same and the two speeds are the same (so the car, continuing to accelerate, will not be passed by the bicyclist).

You have two equations, 8t= (1/2)a^2+ 40, and 8= at, to solve for a and t.

3. ## Re: car and bike

im able to solve them thanks i get a=.8 how are there speeds equal when they meet