Example+3+(homework+problem)



1. At t = 0, is the particle moving to the right or to the left? Explain your answer. Please also give the velocity function that you use!! (Joe).

To determine the direction that the particle is moving in, we take the derivative of the position function to find the velocity. Then when you plug in 0 for t into the velocity equation,we get that the velocity is 9. Since the velocity is positive, this means that the particle is moving in the right direction.

2. At t = 1, is the velocity of the particle increasing or decreasing? Explain your answer. Please also give the acceleration function that you use!! (Han) x'(t)=v(t)=3t^(2)-12t+9 v'(t)=a(t)=6t-12 a(1)=-6 At t=1, the velocity of the particle is decreasing. We know this because at t=1, the derivative of the velocity of the particle (acceleration) is negative, which means that the velocity has a negative slope at t=1. Therefore, the velocity of the particle is decreasing at t=1.

3. Find all values of t for which the particle is moving to the left. Explain your answer in terms of velocity! (Elizabeth)

We know that the particle is moving to the left when v(t) is negative. To find the velocity function take the derivative of the position function. By testing the values 1 and 3 in the velocity equation we can see where the velocity function is negative (moving to the left). Therefore the particle is moving to the left from t=1 to t=3

4. Find the total distance traveled by the particle over the time interval [0, 5]. Explain your answer in terms of position at a particular moment (i.e. when has it moved furthest to the left? How far to the left? Then how far to the right?). Use x(t) for this. I would also like you (Nick Moore) to find the total //displacement// at the end of 5 seconds; how far is the particle from where it initially started?

The formula would be the integral from 0 to 5 of v(t) which would just end up being x(t) from (0,5). x(t) = t^3 - 6t^2 + 9t + 11 x(0) = 11 x(1) = 15 x(2) = 13 x(3) = 11 x(4) = 15 x(5) = 31 Total distance = 11 + 15 + 13 + 11 + 15 + 31 = 96 Farthest to the left x(3) and x(0) = 11 Farthest to the right x(5) = 31 Total displacement = x(5) - x(0) = 20