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embizze | High School Teacher | (Level 1) Educator Emeritus

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We are given the acceleration function `a(t)=t-1.5 ` on the domain `0 <= t <= 3.25 ` . We are also given `v(1)=.4 ` , and the total time as 3.25 minutes.

** To find the velocity function we integrate the acceleration function with respect to `t ` :

`v(t)=int (t-1.5)dt= t^2/2-3/2t+c `

Since v(1)=.4 we can find the constant:

`1/2-3/2+c=.4 ==> c=1.4 `

** To find a formula for the distance traveled we integrate the velocity function with respect to ` t ` :

`s(t)=int (t^2/2-3/2t+1.4)dt=t^3/6-t^2/4+1.4t+C `

(1) The velocity function is `v(t)=t^2/2-3/2t+1.4 ` in `"miles"/"minute" ` with `0<=t<=3.25 `

The distance function is `s(t)=t^3/6-t^2/4+1.4t ` in miles for `0<=t<=3.25 ` . (Since the initial distance traveled, at t=0, is zero the constant of integration is zero.)

(2) The initial velocity is `v(0)=1.4 "miles"/"min" ` or 84mph.

The final velocity is `v(3.25)=1.8063 "miles"/"min"~~108.4 ` mph.

The minimum velocity occurs at t=1.5 and `v(1.5)=.275"miles"/"min" ~~16.5 ` mph.

** We can compute the minimum velocity by taking the derivative of the velocity function, which is just the acceleration function, and setting it equal to zero as minimums occur when the derivative is zero or fails to exist. **

(3) Assuming that the velocity is minimum when we crest the hill, the time to the top of the hill is 1.5 minutes.

(4) The total distance is given by `s(3.25)~~7.6307 ` miles.

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