Question

Sorry this is really long. Just wondering how I would do each of these

A particle is moving with velocity v(t) = t^2 – 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.

The average velocity over the interval 0 to 8 seconds
The instantaneous velocity and speed at time 5 secs

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

The time interval(s) when the particle is
going faster
slowing down

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

Answer 1

remember that d(s)/dt = v(t)
and acceleration = v ' (t)

s(t) = (1/3)t^3 - (9/2)t^2 + 18t + c
given s(0) = 1
1 = 0-0+0+c
c = 1

s(t) = (1/3)t^3 - (9/2)t^2 + 18t + c

s(0) = 1
s(8) = 512/3 - 288 + 144+1
= 83/3

avg velocity for first 8 seconds
= (83/3 - 1)/(8-0)
= 10/3 m/s
v(5) = 25 - 45 + 18 = -2 m/s

the particle is moving to the right if v > 0
t^2 – 9t + 18>0
(t-6)(t-3) > 0
So v > 0 for 3 < t < 6

a(t) = 2t - 9

the object is moving faster when a > 0
and slower when a < 0

take over