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At the instant the traffic light turns green, Stan Speedy starts from rest and a

ID: 1327373 • Letter: A

Question

At the instant the traffic light turns green, Stan Speedy starts from rest and accelerates at "a". At the same instant Kathy Kool starts from rest "d" behind Stan and accelerates at "5/4a". Given [a,d]

a. How far beyond Stan's starting point do they meet again?    Ans. 4d

b. The time before they meet again.     Ans. sqrt(8d/a)

c. How fast is each moving when they meet again.

d. Plot the position as a function of time for both. Take x = 0 at Kathy's initial point.

I am suppose to get those answers for a. & b. but I don't know how. Can someone show me and please explain their reasoning for the important steps for me?

Explanation / Answer

a)

They meet first time when S = s + d;

where S = 1/2 AT^2 and s = 1/2 aT^2

T when they meet T^2 (1/2)(A - a) = d

Then s = 1/2 a 2d/(1/4)a = 4d

b)

The time before they meet again

4d = 1/2 aT^2; so T = sqrt(8d/a)

c)

t = sqrt(8*d / a)

v = a * t

v = sqrt( 8 * d / a) * a

v = sqrt( 8 *d * a)