A bus accelerates at 2.4 m/s2 from rest for 10 s. It then travels at constant speed for 25 s, after which it slows to a stop with an acceleration of magnitude 1.5 m/s2.

Question

A bus accelerates at 2.4 m/s2 from rest for 10 s. It then travels at constant speed for 25 s, after which it slows to a stop with an acceleration of magnitude 1.5 m/s2.
(a) What is the total distance that the bus travels?
(b) What was its average velocity?
If possible, could you explain things and how you got the formulas too? Thank you!

Henry · Answer

V1 = a*T1 = 2.4 * 10 = 24 m/s.

d1 = 0.5a*T1^2 = 1.2*10^2 = 120 m.
a. d2 = V1*T2 = 24 * 25 = 600 m.

V^2 = V1^2 + 2a*d3 = 0
d3 = -V1^2/2a = -(24^2)/-3 = 192 m.

d = d1+d2+d3 = 120 + 600 + 192 = 912 m. = Total distance traveled.

b. V = V1 + a*T3 = 0.
T3 = -V1/a = -24/-1.5 = 16 s.

T = T1+T2+T3 = 10 + 25 + 16 = 51 s.

V = d/T = 912m/51s. = 17.9 m/s. = Avg.
velocity.