Question

1. The velocity of an object moving along a linear path is defined v(t) = t 3 - 3t2, where v is measured in meters per second and t ≥ 0. If the object’s position at t = l is known to be 5 meters from the origin, answer the following.
Find the object’s position at t = 3.
Find the total distance traveled by the object on the interval 0 ≤ t ≤ 5.

I solved the first part, but I can't get the second part? My teacher said i need to split my integral ∫0 5 t^3 - 3t^2 dt into two integrals because part of it is where the velocity is negative.

Answer 1

This was my answer:

a. Integrate V(t) which is (1/4)t^4 - t^3 + c. Plug in t=1 to find C because we know it has to equal 5. C = 23/4. Now to answer what is asked, plug in t=3. (1/3)(3^4) - (3^3) + 23/4. The object's position is -1.

b. Given an interval from 0 to 5, we have to use a definite integral using V(t) to find the total distance. ∫05 t^3-3t^2dt. This becomes (1/4)t^4-t^3 + C. We know C is 23/4 so we can simply use (1/4)5^4-5^3+23/4 = 37.

Answer 2

I see you closed the question, either way, because we have a definite integral we do not need a constant of integration. Secondly, the question is asking for the distance, not displacement, that means that we will need a second integral. It appears that x=3 is the zero of the graph, so the integral for distance would look like \[-\int\limits_{0}^{3} v(t) + \int\limits_{3}^{5} v(t)\]

Answer 3

would thre answer be 31.25 then?

Answer 4

6.75+38=44.75, it looks like you neglected the negative preceding the first integral. Distance is always positive, if you prefer it could be ∫03 absolutevalue(v(t))+∫35v(t).