yes, v = 6 t - 3 t^2
dv/dt = 0 at max or min = 6 - 6 t
so max or min at t = 1
max or min?
d^2v/dt^2 = -6 so max
what is that max v at t = 1
v(1) = 6-3 = 3
so C.
A particle is moving along the x-axis according to the position function s(t)=3t^2-t^3. What is its maximum velocity?
A. 1
B. 2
C. 3
D. 4
E. none of these
I thought velocity was the derivative but I'm not sure with this problem.
2 answers
Take the first derivative, then set that equal to zero, then solve you should get t=0 and t=6.
We know that in order for their to be a maximum, our graph must be concave down. Which means that our objects goes up then goes down. With that our velocity is equal to zero at the starting position and when we have a horizontal tangent which means the maximum y value and our final position. So if the object takes 6 seconds to fall back down then it takes 3 seconds to to go up and three seconds to come down. So after three seconds our object is at its maximum.
We know that in order for their to be a maximum, our graph must be concave down. Which means that our objects goes up then goes down. With that our velocity is equal to zero at the starting position and when we have a horizontal tangent which means the maximum y value and our final position. So if the object takes 6 seconds to fall back down then it takes 3 seconds to to go up and three seconds to come down. So after three seconds our object is at its maximum.