Since the deceleration is constant, we know that:
a(t) = C, for some constant C.
Integrating the acceleration function gives the velocity function to be:
v(t) = Ct + D, where D is another constant.
If we take t = 0 to be when the car is slowing down, then:
v(0) = 15 ==> D = 15.
v(t) = Ct + 15.
Integrating the velocity function gives the position function to be:
s(t) = (1/2)Ct^2 + 15t + E, where E is a third constant.
If we take the initial position of the car to be 0m, then:
s(0) = 0 ==> E = 0,
s(t) = (1/2)Ct^2 + 15t.
(a) We want the value of C knowing that the car stops after traveling 45 m.
The car stops when:
Ct + 15 = 0 ==> t = -15/C.
Plugging this into s(t) gives the position of the car in terms of C to be:
s(-15/C) = (1/2)(C)(-15/C)^2 + 15(-15/C)
Then, the car travels 45 m when s(t) = 45, so we want:
-225/(2C) = 45 ==> C = -2.5 m/s^2.
(b) The car is traveling 3 m/s when:
3 = -2.5t + 15 ==> t = 8/3.
Plugging t = 8/3 back into the position function yields the required distance to be:
s(8/3) = (1/2)(-2.5)(8/3)^2 + 15(8/3) ≈ 31 m.
(c) With v(t) = Ct + 15 = -2.5t + 15, the car stops when:
-2.5t + 15 = 0 ==> t = 6 s.
· 8 years ago