A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. what is the distance covered before the car comes to a stop? (round your answer to one decimal place.)
Answer:
p[0] = 0
v[0] = 50 mi/h = 50 * 5280 / 3600 ft/s = 50 * (22/15) ft/s = 10 * 22/3 ft/s = 220/3 ft/s
a = -50 ft/s^2
a = -50
Integrate to find velocity
v = -50t + C
220/3 = -50 * 0 + C
220/3 = C
v = 220/3 - 50t
Find when v = 0
0 = 220/3 - 50 * t
50 * t = 220/3
t = 220 / 150
t = 22/15
Integrate the velocity function to find the position function
p = (220/3) * t - 25 * t^2 + C
0 = (220/3) * 0 - 25 * 0^2 + C
0 = C
p = (220/3) * t - 25 * t^2
t = 22/15
p = (220/3) * (22/15) - 25 * (484/225)
p = (4840 / 45) - 484 / 9
p = 484 * (10/45 - 1/9)
p = 484 * (2/9 - 1/9)
p = 484 * (1/9)
p = 484/9
p = 53.777777777777777777777777777778
53.8 feet
