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
