john is riding a bike along a straight path. For 0 <= t <= 20, John's velocity is given by a differentiable function v.

Question

Selected values of v(t), where t is measured in minutes and v(t) is measured in meters per minute, are given in the table below.

t(minutes). 0. 3. 6. 10. 12. 20

v(t) (meters per minute). 0 200. 500. 600. -430. 360

1. Use the data table to estimate the value of v(8).
2. Find John's average acceleration ove the interval 6 <= t <= 12.
3. Sam is jogging along the same path. For 0 <= t <= 5. Sam's velocity is modeled by v(t) = 0.5t^3 - 2t^2 + 100. Find her acceleration at time t = 3.
4. At time t = 3, John is at a position xj = 200 meters and Sam is at a position xs = 300 meters, as measured from Sam's starting point. Find the rate , in meters per minute, at which the distance between John and Sam is changing at time t = 3.

Answer 1

p1. v(8) = (600 - 500) / (10 - 6) = 25 mmin^-2, thus at that rate, in two minutes from t = 6, he gains 50 meters.

therefore v(8) = 550 mmin^-1

p2. Final - Initial velocity over Final - Initial time

a_avg = (-430 - 500) / (12 - 6) = - 155 mmin^-2

p3. Take derivative of velocity function, and use the function to calculate acc at t = 3

dv(t) / dt = a = 1.5t² + 4t

a(t=3) = 25.5mmin^-2

p4. Δx/Δt = 300 - 200 / 3 - 0 = 100 / 3 = 33.33 mmin^-1