the velocity of a dragster t seconds after leaving the starting line is v(t)= 100e^-0.2t  ft/sec...1) estimate the distance covered by the dragster in the first 10 sec if its run. Use 4...

the velocity of a dragster t seconds after leaving the starting line is v(t)= 100e^-0.2t  ft/sec...

1) 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.

2) find the distance travelled exactly

(* how do you find the rectangles? thanks!)

Asked on by zita1234

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beckden | High School Teacher | (Level 1) Educator

Posted on

To find the intervals take the total time (10ssec) and divide by 4 to get 2.5sec

To find the midpoint we take half that to get 1.25sec

So we are going to take the velocity at 1.25sec, 3.75sec, 6.25 sec, and 8.75 multiply by the interval length (2.5secs) and use that to get the velocity.

v(1.25)=100e^(-0.2(1.25))=77.880078ft/sec

v(3.75)=100e^(-0.2(3.75))=47.236655ft/sec

v(6.25)=100e^(-0.2(6.25))=28.650480ft/sec

v(8.75)=100e^(-0.2(8.75))=17.377394ft/sec

We would multiply the intervals by the velocity and then add, but since the interval is fixed we can just add the velocities and then multiply by the interval(2.5sec).

Total of Velocities = 171.14461ft/sec

171.14461∗2.5= 427.86153 ft.

To find it exactly we integrate the velocity from 0 to 10

`s(t) = int v(t) dt = int_0^10 100e^(-0.2t)dt = 100/(-0.2)e^(-0.2t)|_0^10`

`s(t) = -500e^(-0.2t)|_0^10=-500e^(-0.2(10))-(-5e^(-0.2)(0))=`

`s(t)=-500e^(-2)+500e^(0)=500-500e^(-2)=`432.33236ft

Which is very close to our estimate.

So the answers are a) 427.86153 ft. and b) `500(1-e^2)=432.332 ft`

 

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