Question

The acceleration a(t) of a body moving in a straight line is a (t)=8-6t. If the velocity of the body is 25 at t=1 and if s(t) is the distance of the body from the origin at time t, what is s(4)-s(2)?

Answer 1

The distance travelled between the seconds 2 and 4 is 28m.
`s(4) - s(2) = 28m`

Explanation:

The acceleration at any time is modelled by `a(t)=-6t+8`.

First we have to determine the velocity at all times, we can start doing this by integrating the acceleration to get velocity.

`v(t) = inta(t)dt = -3t^2+8t+c`

The constant c can be found using the given information that `v(1) = 25`

So `v(1) = 25=-3(1)^2+8(1)+c`
`c = 20`

`v(t) = -3t^2+8t+20`

Distance can be found by integrating the velocity function. Knowing that we are trying to find the distance travelled between 2 and 4 seconds we can use a definite integral of the velocity function to find distance.

`s(4) - s(2) = int_2^4v(t)dt`
`= (-(4)^3+4(4)^2+20(4))-(-(2)^2+4(2)^2+20(2))`
`=80-52=28`

The distance travelled between the seconds 2 and 4 is 28m.
`s(4) - s(2) = 28m`