Question

AP PHYSICS SYSTEM OF EQUATIONS

john and ryan are walking towards each other. john begins walking from the starting line of the 100m race at a speed of 2.1m/s. ryan starts walking 2 seconds after john starts. ryan begins walking from the finish line of the 100m race at a soeed of 1.8m/s.
a- how far from the starting line do the students meet?
b-how many seconds did it take them to meet?

i have the answers, just having troubles figuring out how to set this equation up. please show your work!! dont take advantage of the points either please

Answer 1

Walking at a speed of 2.1 m/s, in the first 2 s John would have walked

(2.1 m/s) (2 s) = 4.2 m

Take this point in time to be the starting point. Then John's distance from the starting line at time t after the first 2 s is

J(t) = 4.2 m + (2.1 m/s) t

while Ryan's position is

R(t) = 100 m - (1.8 m/s) t

where Ryan's velocity is negative because he is moving in the opposite direction.

(b) Solve for the time when they meet. This happens when J(t) = R(t) :

4.2 m + (2.1 m/s) t = 100 m - (1.8 m/s) t

(2.1 m/s) t + (1.8 m/s) t = 100 m - 4.2 m

(3.9 m/s) t = 95.8 m

t = (95.8 m) / (3.9 m/s) ≈ 24.6 s

(a) Evaluate either J(t) or R(t) at the time from part (b).

J (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m