Applications of derivatives: Velocity.

Question

The movement of a particle from time t = 0 to t = 1 is given by the equation s(t) = -4t² -8t -7

What is the average velocity of the particle between t = 0 and t = 1?

What is the instantaneous velocity of the particle at t = 1?

What is the acceleration of the particle at t = 1?

Answer 1

v = ds/dt therefore v(t) = s'(t) = -8t - 8

The average velocity between t = 0 and t = 1 is (v(1) + v(0))/2, Using v(t) = -8t - 8 then v(1) = -16 and v(0) = -8 so the average is (-16 + -8)/2 = (-24)/2 = -12 (units not given)

The instantaneous velocity at t = 1 is v(1) = -16

a = dv/dt = v'(t) = -8 so no matter what the time is, the acceleration is always -8