Question

17. The velocity, in meters per second, of a runner is given by the function v(t)=11 et +5t. A. Find the acceleration of the runner at time t = 3 seconds. Include the proper units. B. Find the position function given that the runner starts at a distance of 6 meters.​

Answer 1

a) a= 5 m/s^2

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

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