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.

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

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.

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

shannonytuiot · Answer

@ganeshie8 Can you help me out?

trojanpoem · Answer

$$v_{av} = \frac{ s(8) - s(0) }{ t_{2} - t_{1} }   $$

Instantaneous velocity:

Get the derivative and let t = 5 ( v(5)  )

When the particle is moving right s(t) > 0 (You will get 2 inequalities combine them)

Get the second derivative when a > 0 (going faster) , a < 0 (Slowing down)

replace a with the time function. 

Find distance in left direction + In right direction - s(0)

shannonytuiot · Answer

@TrojanPoem Thank you so much!!!