Question

Question #473da

Answer 1

Non-instantaneous velocity = `v=3.5`

Distance traveled = `d=14.83`

Explanation:

So we are given the equation:

`a(t)=2t+3` (this is acceleration with respect to time)

and the value

`v(0)=-4` (this is the velocity of the object at time 0)

so first we need to find the equation for velocity with respect to time.

So we integrate our acceleration equation to find velocity.

`v(x)=int_0^xa(t)dt`

`v(x)=int_0^x2t+3dt`

`v(x)=x^2 +3x+c`

So since we know what `v(0)` is equal to we can solve for c

`-4=0^2+3(0)=c`
`-4=c`

`v(x)=x^2+3x-4`

(I used x before just so we could visualise where the numbers were coming from) We can now substitute t back in.

`v(t)=t^2+3t-4`

To find displacement over a given time we need to integrate the velocity equation with the range of time we are given.

`0 <= t <= 3` (t is between 0 and 3)

`x=int_0^3t^2+3t-4dt`
`x=21/2`
`x=10.5`

As non instantaneous velocity `=(displacement)/(time)`

`v=10.5/3`

`v=3.5`

Now let's find the total distance travelled which is different from displacement.

`d=int_0^3abs(t^2+3t-4)dt`

The absolute values takes the negative portions of the velocity graph and make them positive so that we are no longer subtracting them in our integral thus giving us total distance and not displacement.

`d=int_0^3abs(t^2+3t-4)dt`

`d=89/6`

`d=14.83`

Hope this answers your question if something is unclear feel free to ask.