We now use the formula to find the distance travelled by the car.

In this case, we have:

`x(t)  = 20+0.2t^3`, so `(dx)/(dt)=0.6t^2`

`y(t)  = 20t − 2t^2` giving `(dy)/(dt)=20-4t`

Our lower and upper limits for this example are `t = 0` to `t = 8`.

Substituting these into the distance formula gives:

`text[length]=r` `=int_0^8 sqrt[(0.6t^2)^2+(20-4t)^2]`

Using a computer algebra system (see the answer in Wolfram|Alpha) gives us the length `144.7\ "m"`.

Our answer is reasonable and is consistent with our earlier estimate.