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An object is moving with a position function s(t) = t^3 +6t^2 +7 (meters)
Find the object's velocity and acceleration after 3 seconds

Respuesta :

The velocity and acceleration of the object after 3 seconds are;

Velocity = 63 m/s

Acceleration = 30 m/s²

Distance Position Function

We are given the function that represents the position of an object as;

s(t) = t³ + 6t² + 7 (meters)

  • To get the velocity function, we will differentiate the distance function above to get;

v(t) = s'(t) = 3t² + 12t

  • Also, to get the acceleration function, we will differentiate the velocity function to get;

a(t) = v'(t) = 6t + 12

Thus;

v(3) = 3(3²) + 12(3)

v(3) = 63 m/s

Likewise;

a(3) = 6(3) + 12

a(3) = 30 m/s²

Read more about distance position function at; https://brainly.com/question/24786395

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