AP Calculus AB MCQ

Question

A particle moves along the x-axis so that at any time t ≥ 0, its velocity is given by v(t) = cos(2 - t^2). The position of the particle is 3 at time t = 0. What is the position of the particle when its velocity is first equal to 0?

a. 0.411

b. 1.310

c. 2.816

d. 3.901

e. 3.411

Answer 1

The velocity will be zero when cos (2-t²) = 0, ie when 2-t² = π/2, 3π/2 ...

First solution 2-t² = (π/2) so t = 0.6551

To find the position integrate v(t) from t = 0 to t = 0.6551 then add 3 to this (as particle was at position 3 at t=0).

which gives (c).

You can integrate on the TI-84 (MATH 9), make sure calculator is set to radians