At what time t does the particle change direction?

Question

An object moves in a straight line with acceleration a(t) = t^2 ft/s^2. If the initial velocity of the object is −9 ft/sec, at what time t does the particle change direction?

Answer 1

Hi Ana,

First, let's consider what the question is asking. When exactly does the particle change directions? Well, that'll occur when the velocity changes from negative to positive or vice versa. This can only happen when v(t) = 0. So, now we have our goal, we need to find all of the times, t, such that v(t) = 0 and then check that the sign of v(t) changes.

To find when v(t) =0, we need an equation for v(t). Thus, we can relate velocity and acceleration by noting that the anti derivative of acceleration is velocity. Thus we have:

v(t) = int( a(t) ) where int stands for the anti derivative

v(t) = int (t^2)

v(t) = 1/3 t^3 + c

Now, before we can set v(t) = 0, we need to solve for our constant of integration, c. To do so, we can use the last piece of given information. We know that initially, that is at time t=0, the velocity of the object is -9 ft/sec. So we have:

-9 = v(0) = 1/3 (0^3) + c = c

So c = -9 giving us v(t) = 1/3 t^3 -9.

All that is left is to find times t such that v(t) =0. So we solve

1/3 t^3 -9 = 0

t^3 - 27 = 0 (multiply everything by 3)

(t-3) (t^2+3t+9) = 0 (the difference of two cubes)

Since t^2+3t+9 is always positive, our only real answer is t=3. And indeed, we can check that the velocity is negative when t<3 and positive when t>3.

Thus the particle changes directions exactly once at t=3 seconds.

Hope my explanation helps,

Philip

Answer 2

The acceleration is the first derivative of the velocity. Hence, to get the velocity from the acceleration requires integration in time.

v = ∫a dt

v = ∫t² dt

v = (1/3)t³ + v_o where v_o is the arbitrary constant.

v(t = 0) = v_o = -9 ft/sec

v = (1/3)t³ - 9

The acceleration is positive and the initial velocity is negative. Hence, the velocity will becoome less and less negative, go through zero and then become positive. The object switches direction when v = 0.

0 = (1/3)t³ - 9

(1/3)t³ = 9

t³ = 27

Take the cube root of each side.

t = 3 sec