OpenStudy (anonymous):

At a distance of 45m from a traffic light, a car traveling 15m/sec is brought to a stop at a constant deceleration. a) What is the value of deceleration b) How far has the car moved when its speed has been reduced to 3m/sec? c) How many seconds would the car take to come to a full stop?

4 years ago
OpenStudy (anonymous):

Sadly I'm stuck... a. I got this equation: \[15t-at^2=45\] Then I was trying to find a way to solve for a by taking a derivative but came up short. What solution did you get? I suppose it might help to know t=3+s

4 years ago
OpenStudy (anonymous):

i don't even know how to start this. i'm very confused, how did you come up with that equation?

4 years ago
OpenStudy (anonymous):

In that equation I was attempting to relate distance traveled to speed. So 15t= distance traveled over a time, and at^2=deceleration over a time. The sum of the two will equal 45.

4 years ago
OpenStudy (anonymous):

i'll keep at this for 15 more minutes since no one else has come to help to see if i can get an answer.

4 years ago
OpenStudy (anonymous):

thank you : )

4 years ago
OpenStudy (anonymous):

so you made up that equation by yourself? its not from a formula or anything?

4 years ago
OpenStudy (anonymous):

Yes, sort of. Alright I don't want to waste more time on this, so i found this: 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. This gives: 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, and: 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) = -225/(2C) 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. http://ca.answers.yahoo.com/question/index?qid=20110917121026AAkNqPh If you haven't already googled the question. I can sort of explain which each step means.

4 years ago
OpenStudy (anonymous):

i wouldn't mind an explanation of each step.

4 years ago