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

1. The instantaneous velocity and speed at time 5 secs

1. The time interval(s) when the particle is moving right

1. The time interval(s) when the particle is
2. going faster
3. slowing down

1. Find the total distance the particle has traveled between 0 and 8 seconds

Answer 1

Basics of particle motion:

s(t) is position

s '(t) = v(t) is velocity

s ''(t) = v '(t) = a(t) is acceleration

If given velocity, as in this problem, must

1. take its derivative to find acceleration (speeding up or slowing down)
2. integrate velocity to find position function.

So s(t) = ∫ (t²-9t+18) dt = t³/3 - 9t²/2 + 18t + C

Find C by using s(0)=1

0³/3 - 9(0²)/2 + 18(0) + C = 1, so C = 1

Now our final s(t) = (1/3)t³ - (9/2)t² + 18t + 1

We already know v(t) = t² - 9t + 18 [ note that his factors into (t-3)(t-6) ]

And a(t) = 2t - 9

Please watch video now for how to use the above information to solve the problems.