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) average velocity = (v(8) - v(0))/(8 - 0) = (10 - 18)/8 = -1m/s

 

2) v(5) = 5^2 - 9(5) + 18 = 25 - 45 + 18 = -2m/s

 

3) set v(t) = 0

 

t^2 - 9t + 18 = 0

 

(t - 6)(t - 3) = 0

 

t = 6 t = 3

 

When 0<t<3, the velocity function is positive which means the particle is moving to the right. Also, when 6<t<8 the particle is moving right.

 

4) find the acceleration function

 

a(t) = 2t - 9

 

2t - 9 = 0

 

2t = 9

 

t = 9/2

 

When 0<t<9/2, the particle is slowing down since the acceleration is getting closer to zero. When 9/2<t<8, the particle is speeding up because the acceleration is increasing away from zero.

 

5) Int(t^2 -9t + 18) = (t^3)/3 - (9t^2)/2 + 18t + c

 

s(0) = 1

 

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

 

Now we use the 3 intervals in part 3 to find the total distance

 

s(3) - s(0) = (9 - 81/2 + 54 + 1) - 1 = 45/2

s(6) - s(3) = ( 72 -162 + 108 + 1) - 47/2 = -9/2

s(8) - s(6) = (512/3 - 288 + 144 + 1) - 19 = 26/3

 

All distances are scalar quantities so they are positive

 

45/2 + 9/2 + 26/3 = 107/3 meters