Find the displacement and the distance traveled by the particle

Question

The velocity function v(t)= -t^2+6t-8 is  for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-2,6]

Answer 1

Hi Shai M.,

So, what is the calculus relationship between displacement, velocity, and acceleration?

If you know that velocity is d(displacement)/dt , you are where you need to be.

And you know what to do to, to reverse a derivative? You integrate!

So integrate (-t^2+6t-8) over the stated limits, that will give you the displacement.

However, it won't give you the distance, necessarily -- if there is a reversal of the sign of the velocity, over that interval. So take the NEGATIVE of the velocity expression (for convenience), that makes x^2 - 6x +8 . Factor that: (t - 2)(t-4) . That should tell you that the velocity flips sign at t=2 and t=4. So to figure the distance, just break that integration interval into 3 parts: [-2, 2], [2, 4], and [4,6]. Calculate each of these separately, then take their absolute values and sum them. That's the distance.

--Cheers, --Mr. d.