An object moves along a coordinate line with velocity v(t) = t(6 − t) m/s. Its initial position is 2 units to the left of the origin.

Question

(a) Find the position of the object 16 seconds later.

(b) Find the total distance traveled by the object during the first 16 seconds.

Answer 1

(a)

Position x(t) is the integral of velocity.

x(t) = ∫t(6-t)dt = ∫6tdt - ∫t²dt = 3t² - t³/3 + C

The constant of integration, C, is determined from the initial condition

x(0) = -2

Therefore C = -2.

x(16) = 3×16² - 16³/3 - 2 = -599.3

(b)

Look at the function v(t) to understand the movement.

At t=0 velocity is 0.

Then velocity is positive, meaning that the object moves in the positive direction,

until t=6 when the objects stops and reverses direction.

Thereafter velocity is always negative, so the object continues in the

negative direction forever.

At time t=6 it reached the point

x(6) = 3×6² - 6³/3 - 2 = 34

Between the times 0 and 6 it traveled the distance x(6)-x(0).

Then between times 6 and 16 it traveled the distance x(6) - x(16).

So it traveled the total distance

2x(6) - x(0) - x(16) = 2×34 + 2 + 599.3 = 669.3