A car traveling initially at +7.0m/s accelerates uniformly at the rate of +0.80 m/s^2 for a distance of 245 m.

a. what is its velocity at the end of the acceleration?
b. what is its velocity after if accelerates for 125 m?
c. what is its velocity after it accelerates for 67 m?

a. what is its velocity at the end of the acceleration?
Let D = distance traveled (245 m), Vi = initial velocity (7.0 m/s), a = acceleration (0.80 m/s^2), t = time

D = (1/2)at^2 + (Vi)t
245 = (0.5)(0.80)(t^2) + 7t
245 = 0.40t^2 + 7t
0 = 0.40t^2 + 7t - 245 now use quadratic equation to find t where 0.40 = a, 7 = b, and -245 = c
t = (-7 +/- sqrt((7)^2 - (4)(0.40)(-245))) / (2*0.40)
t = (-7 +/- sqrt(49 + 392)) / (0.80)
t = (-7 +/- sqrt(441)) /(0.80)
t = (-7 +/- 21) / (0.80)
t =-28 / (0.80) or t = 14 / (0.80)
t = -35 or t = 17.5 a negative time makes no sense so it takes 17.5 seconds to travel 245 meters

Vf = a(t) + Vi
Vf = (0.80)(17.5) + 7.0
Vf = 14.0 + 7.0
Vf = 21 m/s ANSWER

b) what is its velocity after if accelerates for 125 m?
125 = (0.5)(0.80)(t^2) + 7t
125 = (0.5)(0.80)(t^2) + 7t
125 = 0.40t^2 + 7t
0 = 0.40t^2 + 7t - 125 now use quadratic equation to find t where 0.40 = a, 7 = b, and -125 = c
t = 11.0 seconds
V = 15.8 m/s ANSWER

c) what is its velocity after it accelerates for 67 m?
67 = (0.5)(0.80)(t^2) + 7t
67 = (0.5)(0.80)(t^2) + 7t
67 = 0.40t^2 + 7t
0 = 0.40t^2 + 7t - 67 now use quadratic equation to find t where 0.40 = a, 7 = b, and 67 = c
t = 6.9 seconds
V = 12.5 m/s

Gain in velocity after 245m. = sqrt. (2ad), = 19.8m/sec. Add initial velocity, = 26.8m/sec.
Can you do the rest? They're all the same.

Wtf ???