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17. The velocity, in meters per second, of a runner is given by the function [tex]v(t) = 11 e^t + 5t[/tex].

A. Find the acceleration of the runner at time [tex]t = 3[/tex] seconds. Include the proper units.

B. Find the position function given that the runner starts at a distance of 6 meters.

Answer :

Answer:

a) a= 5 m/s^2

b) P(t) = 2.5 t^2 + 11 t + 6

Step-by-step explanation:

The velocity function is V(t) = a*t + Vo

In this case: V(t) = 5t + 11

It means the constant acceleration a = 5 m/s^2

while the initial velocity (at time t=0s) is 11 m/s

The position function is:

P(t) = 1/2 A*t^2 + Vo*t + Po

In this case:

P(t) = 2.5 t^2 + 11 t + 6

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