# AP CALC. AB

1.The position of a particle moving on the line y = 2 is given by x(t)= 2t^3-13t^2+22t-5 where t is time in seconds. When is the particle at rest?
a. t =0.268, 2.500, and 3.732
b. t = 0, 1.153, and 3.180
c. t = 1.153, 2.167 and 3.180
d. t = 2.167
e. t = 1.153 and 3.180
2.A particle moves on the x-axis in such a way that its position at time t is given by x(t) 3t^5-25t^3+60t For what values of t is the particle moving to the left?
a. 1 < t < 2 only
b. t < -2, -1 < t < 1, and t > 2
c. –2 < t < 1 only
d. –1 < t < 1 and t > 2
e. –2 < t < -1 and 1 < t < 2
3. A particle moves along the x-axis so that its position at any time t>0 is given by
x(t)=t^4-10t^3+29t^2-36t+2.For which value of t is the speed the greatest?
a. t = 1
b. t = 4
c. t = 5
d. t = 3
e. t = 2
Please walk me through each step!

1. #1 The particle is at rest when its velocity is zero. So, we need dx/dt=0
dx/dt = 6t^2-26t+22
So, solve for t

#2 moving left ==> velocity is negative
dx/dt = 16t^4-75t^2+60
So, solve for dx/dt < 0

#3 greatest speed is max velocity. That means dx/dt has a max, so d^2/dt^2 = 0
v=x' = 4t^3-30t^2+58t-36
v'=x" = 12t^2-60t+58
So, solve for t in x" = 0
Also, check to make sure you have found a max, not a min. That is, d^2v/dt^2 < 0.

posted by Steve

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