Answer:
a_a = -3.2 ft/s^2 , a_b = 3.723 ft/s^2
t = 5.909 s
Explanation:
Given:
- Initial velocity of A v_i,a = 30 ft /s
- Initial distance s_o = 0
- Length of the exchange zone s_f = 65 ft
- Time taken t = 2.5 s
Start out by A's velocity as he gets to the end of the exchange zone.
Part a
Runner A decelerates at a uniform rate, so you can use the equation:
s_f = s_o + v_i,a*t + 0.5a_a*t^2
65 = 0 + 30*2.5 + 0.5*a_a*2.5^2
a_a = -20 / 2.5^2
a _a= -3.2 ft/s^2
Runner B accelerates at v_f,a as final velocity at a uniform rate, so you can use the equation:
v_f,b^2 - v_i,b^2 = 2*a_b*s
a_b = (30 -3.2*2.5)^2 /2*65
a_b = 3.723 ft/s^2
Part b
When Runner B should begin running:
t = (v_f,b - v_i,b) / a_b
t = (30 - 3.2*2.5) / 3.723
t = 5.909 s
