This is a question about Kinematics and making an equation.

Question

Person A, moving at a constant speed of 4.0 m/s passes person B who is moving at a 1.5 m/s in the same direction. At the instant B is passed, she increases her speed at a constant rate until she catches A. What is the final speed of B?

Answer 1

Starting at the point where A and B are together, A then travels a distance d = 4t in t seconds.  B accelerates and travels a distance d = (1/2)at² + 1.5t in t seconds.  When B catches back up to A, they have traveled the same distance, so set the two distances equal:

 

(1/2)at² + 1.5t = 4t

(1/2)at² - 2.5t = 0

at² - 5t = 0

t(at-5) = 0

 

t = 0 and at = 5

 

So A and B are together at t = 0 when they start out and again when at = 5.  Any combination a and t that produces a product of 5 works, such as a = 1 and t = 5 or a = 2 and t = 2.5,  But it turns out that it doesn't matter, we just need to know the product at.  B's speed when she catches A again is:

 

v = at + 1.5 = 5 + 1.5 = 6.5 m/s

 

So no matter if B accelerates slowly or quickly, she will be going 6.5 m/s when she catches up with A.