# the velocity of a dragster t seconds after leaving the starting line is v(t)= 80te^-0.3t ft/sec...estimate the distance covered by the dragster in the first 10 sec if its run. Use 4 subintervals...

## the velocity of a dragster t seconds after leaving the starting line is v(t)= 80t*e*^-0.3t ft/sec...

estimate the distance covered by the dragster in the first 10 sec if its run. Use 4 subintervals of equal length & use the midpoint of each subinterval to construct rectangles.

Any Help would be greatly appreciated a can't seem to find an exact number any where close to the estimated value

### 1 Answer | Add Yours

You need to remember that distance is the indefinite integral of velocity such that:

`int v(t)dt = s(t)`

`int 80t*e^(-0.3t) dt = s(t)`

You need to use integration by parts such that:

`int udv = uv - int vdu`

You should come up with the substitution `u = 80t ` and `dv = e^(-0.3t)` such that:

`u = 80t=gt du = 80dt`

`dv = e^(-0.3t) =gt v = e^(-0.3t)/(-0.3)`

`int 80t*e^(-0.3t) dt = 80t*e^(-0.3t)/(-0.3) + (80/0.3)int e^(-0.3t)dt `

`int 80t*e^(-0.3t) dt = -800t*e^(-0.3t)/3- 8000e^(-0.3t)/9 + c`

Hence, `s(t) = -800t*e^(-0.3t)/3 - 8000e^(-0.3t)/9 + c`

If t = 10 sec , then `s(10) = 8000*e^(-3)/3+ 8000e^(-3)/9` .

You need to factor out `8000e^(-3)/3` such that:

`s(10) = (8000e^(-3)/3)(1 + 1/3)`

`s(10) = (32000e^(-3)/9) =gt s(10)~~180.640 ft`

**Hence, estimating the distance travelled after 10 sec yields `s(10) ~~180.640 ft` .**