# the acceleration of an object is given by the equation `a(t)=7t^3 - 3t^2 - 4 ` m/s^2. Determine its velocity and position after 5 seconds, assuming s(0)=3 m and v(0)= 1m/s After 5 seconds, the...

the acceleration of an object is given by the equation `a(t)=7t^3 - 3t^2 - 4 ` m/s^2. Determine its velocity and position after 5 seconds, assuming s(0)=3 m and v(0)= 1m/s

After 5 seconds, the object's velocity is= ( )m/s

and its position is=( )m

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Expert Answers

ishpiro | Certified Educator

Velocity is the antiderivative of the acceleration, so

`v(t) = 7 t^4/4 - 3 t^3/3 - 4t + v_0`

Since *v*(0) = 1 m/s, `v_0 = 1` m/s

After 5 seconds (*t* = 5 s)

`v(5) = 7/4 * 5^4 - 5^3 - 4*5 + 1=949.75` m/s

The position is antiderivative of the velocity:

`s(t) = 7/4 * t^5/5 - t^4/4 - 4*t^2/2 + t + s_0`

Since *s*(0) = 3 m, `s_0 = 3` m

`s(t) = 7/20 * t^5 - t^4/4 - 2t^2 + t + 3`

After 5 seconds

`s(5) = 7/20 * 5^5 - 5^4/4 - 2*5^2 + 5 + 3` = 895.5 m

*v*(5) = 949.75 m/s

*s*(5) = 895.5 m