A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive

Question

A particle is moving with velocity v(t) = t² – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero, that is, s(0) = 1.



1. The average velocity over the interval 0 to 8 seconds
2. The instantaneous velocity and speed at time 5 secs
3. The time interval(s) when the particle is moving right
4. The time interval(s) when the particle is
5. going faster
6. slowing down
7. Find the total distance the particle has traveled between 0 and 8 seconds

Answer 1

The distance s(t) is the integral of v(t):

s(t) = t³/3 - 9t²/2 + 18t + C

The constant C is determined from the boundary condition s(0) = 1, so

s(t) = t³/3 - 9t²/2 + 18t + 1

1.

The average velocity is

(s(8) - s(0))/8 =

8²/3 - 9x8/2 + 18 =

3.333

2.

v(5) = 5² - 9x5 + 18 = -2

The instantaneous velocity is -2.

That means the instantaneous speed is 2, and velocity is directed to the left.

3. The particle is moving right, when its velocity is positive.

The velocity is a parabola with roots t=3 and t=6

(which you can calculate by solving t² - 9t + 18 = 0).

Since the leading coefficient, being 1, is positive,

the velocity is positive on the intervals [0,3] and [6,8].

4. I do not know whether "faster" refers to speed or velocity.

That is, when its speed is increasing going backwards, does that count as "faster"?

Therefore I will answer the question both ways.

The particle is going faster or slower when its speed/velocity is increasing

or decreasing respectively.

To determine that, notice that the parabola representing the velocity

has its vertex at t = 4.5.

The particle velocity is decreasing (it is slowing down) on the interval [0, 4.5)

and it is increasing on the interval (4.5, 8].

The speed, being the absolute value of velocity,

is decreasing on the intervals [0, 3) and (4.5, 6), and

is increasing on the intervals (3, 4.5) and (6, 8].

5. Again I am not sure how "distance traveled" is measured.

That is, when the particle is going backwards is it making a positive or negative

contribution to its distance travelled?

If the answer is "negative" then the solution to 5. is simply s(8) - s(0) = 26.666

Now I assume that the answer is "positive".

The particle is going forward on the intervals [0,3) and (6,8], while going backwards

on the interval (3,6).

So the total distance travelled is

s(3) - s(0) + s(3) - s(6) + s(8) - s(6) =

s(8) - s(0) + 2(s(3) - s(6)) = 35.666