An object's movement has a velocity given by v(t) = t^2-5t+5
A) What is the position function for the particle at any time t≥0?
For this section I calculated the anti derivative, which is \frac{1}{3}t^3-\frac{5}{2}t^2+5t+C (At least I think it is) And I'm not sure if that finishes the question or if I have to somehow calculate C... but I'm not sure how to do that.
B) When does the particle change directions?
C) Find the displacement and total distance traveled from time t = 0 to t = 5.
3 answers
The anti derivative i got is 1/3t^3 - 5/2t^2 + 5t + C
1/3t^3 - 5/2t^2 + 5t + C is ok
let's call it s(t)
s(t) = 1/3t^3 - 5/2t^2 + 5t + C
you want the distance between t = 0 and t = 5
s(0) = 0-0+0 + C
s(5) = (1/3)(125) - (5/2)(25) + 5(5) + C
= 25/6 + C
distance covered between the two times = 25/6 + C - C = 25/6 <---- part c
b) it would change direction when v ' (t) = 0
v(t) = a(t) = 2t - 5 , where a(t) is the acceleration
2t - 5 = 0
t = 2.5
let's call it s(t)
s(t) = 1/3t^3 - 5/2t^2 + 5t + C
you want the distance between t = 0 and t = 5
s(0) = 0-0+0 + C
s(5) = (1/3)(125) - (5/2)(25) + 5(5) + C
= 25/6 + C
distance covered between the two times = 25/6 + C - C = 25/6 <---- part c
b) it would change direction when v ' (t) = 0
v(t) = a(t) = 2t - 5 , where a(t) is the acceleration
2t - 5 = 0
t = 2.5
it would change direction when v(t) = 0 and v'(t) ≠ 0
That is, it stops and then reverses direction.
That is, it stops and then reverses direction.