Calculus velocity question..

Question

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.

 

What is ...

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 going faster slowing down

 

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

Answer 1

1.

The average velocity = s(t2)-s(t1)/(t2-t1) while t2>t1. So t2=8, t1=0.

But how to find s(t)? (Hint: s'(t)=v(t)=t^2-9t+18 with the initial condition s(0)=1.)

Then Plug Everything In!

 

2.

Just Plug t=5 to v(t).

 

3.

Find all the ts that satisfy v(t)>0.

 

4.

When the particle goes faster, v(t) should become bigger, with the acceleration function a(t)=v'(t), in this case, a(t)>0.

 

When it slows down, a(t)<0.

 

5.

Integrate the function with the upper limit t=8, lower limit t=0.

s(8) - s(0) = ∫[u=0 to 8] v(u) du

(However, this only for the absolute distance it has travel, which is there may be offsets between positive travel and negative travel)

 

The total distance, without regarding to the direction, should be divided into parts and take absolute value for those time interval where the distance is in minus sign.