Is this correct?
A point moves along the y axes (marked in feet) so that its position at time t (in seconds) is given by f(t)=t^3 - 9t^2 + 15t. Your answer should include the appropriate units.

(a) Find the position of the particle at t = 2 seconds.
f(2) = 2^3 -9(2)^2 + 15(2) = 2 ft.

(b) Find the velocity of the point at time t = 2 seconds.

v=f'(t) = 3t^2 - 18t + 15
f'(2) = 3(2)^2 - 18(2) + 15 = -9 ft./sec.

(c) Find the times at which the velocity is zero.
3t^2 - 18t + 15 = 0
3 (t^2 - 6t + 5)
3 (t-5)(t-1) = 0
t = 5, t = 1

The times at which the velocity is zero are at 5

Question

Is this correct? 
A point moves along the y axes (marked in feet) so that its position at time t (in seconds) is given by f(t)=t^3 - 9t^2 + 15t. Your answer should include the appropriate units.

(a) Find the position of the particle at t = 2 seconds.
f(2) = 2^3 -9(2)^2 + 15(2) = 2 ft.

(b) Find the velocity of the point at time t = 2 seconds. 

v=f'(t) = 3t^2 - 18t + 15
f'(2) = 3(2)^2 - 18(2) + 15 = -9 ft./sec.

(c) Find the times at which the velocity is zero.
3t^2 - 18t + 15 = 0
3 (t^2 - 6t + 5)
3 (t-5)(t-1) = 0
t = 5, t = 1

The times at which the velocity is zero are at 5

jim_thompson5910 · Answer

all 3 answers are good

anonymous · Answer

at 5 sec and 1 sec the velocity is zero

jim_thompson5910 · Answer

good catch, don't forget the units

jim_thompson5910 · Answer

and you might want to reorder, but it's not a big deal really

anonymous · Answer

ok thanks.