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.

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.

Well, in this case, velocity is indeed the derivative of the position function. So, let's find the derivative of s(t) to get the velocity function.

s(t) = 3t^2 - t^3

To find the velocity function, we need to differentiate s(t) with respect to t:

v(t) = d/dt (3t^2 - t^3)

Using the power rule, we can differentiate each term separately:

v(t) = 2 * 3t^(2-1) - 3 * t^(3-1)

v(t) = 6t - 3t^2

Now that we have the velocity function, we need to find the maximum velocity. To do that, we can set the derivative of the velocity function equal to zero:

v'(t) = 6 - 6t = 0

Solving for t, we get:

6t = 6

t = 1

So, at t = 1, the velocity is at a maximum. Now we need to find the value of the velocity at t = 1:

v(1) = 6(1) - 3(1^2) = 6 - 3 = 3

Therefore, the maximum velocity is 3.

Hence, the correct answer is C. 3.

You are correct that velocity is the derivative of the position function. In this case, we can find the maximum velocity by finding the derivative of the position function and then finding the point where the derivative is equal to zero.

To find the derivative of the position function, we can use the power rule. For a function of the form f(t) = at^n, the derivative is f'(t) = nat^(n-1).

Applying the power rule to the position function s(t) = 3t^2 - t^3, we get:

s'(t) = 2(3)t^(2-1) - 3(1)t^(3-1)
= 6t - 3t^2

To find the maximum velocity, we need to find the point where the derivative is equal to zero. So, we set s'(t) = 0 and solve for t:

6t - 3t^2 = 0
3t(2 - t) = 0

This equation is satisfied when either 3t = 0 or 2 - t = 0.

From 3t = 0, we find that t = 0.
From 2 - t = 0, we find that t = 2.

To determine which value of t corresponds to the maximum velocity, we need to examine the sign of the derivative on either side of these points.

For t < 0, s'(t) = 6t - 3t^2 < 0.
For 0 < t < 2, s'(t) = 6t - 3t^2 > 0.
For t > 2, s'(t) = 6t - 3t^2 < 0.

Since the derivative changes sign at t = 0 and t = 2, we can conclude that the maximum velocity occurs at t = 2.

To find the maximum velocity, we substitute t = 2 into the derivative:

s'(2) = 6(2) - 3(2)^2
= 12 - 12
= 0

Therefore, the maximum velocity is 0. So the correct answer is E. None of these.

You are on the right track! Velocity is indeed the derivative of the position function. To find the maximum velocity, we need to find the derivative of the position function with respect to time.

Given the position function, s(t) = 3t^2 - t^3, we can find the velocity function, v(t), by taking the derivative of s(t) with respect to t.

Let's find the derivative of s(t):

s'(t) = d/dt (3t^2 - t^3)

To differentiate the position function, we apply the power rule and the chain rule:

s'(t) = 2(3t^2)(dt/dt) - 3t^2(dt/dt)

Since dt/dt is equal to 1, we simplify further:

s'(t) = 6t - 3t^2

Now that we have the velocity function, v(t) = 6t - 3t^2, we can find the critical points by setting v'(t) = 0:

v'(t) = d/dt (6t - 3t^2)

Applying the power rule:

v'(t) = 6 - 6t

Setting v'(t) equal to 0 and solving for t:

6 - 6t = 0
6t = 6
t = 1

Now that we have the critical point t = 1, we can substitute it back into the velocity function to find the corresponding velocity:

v(t) = 6t - 3t^2

v(1) = 6(1) - 3(1)^2
v(1) = 6 - 3
v(1) = 3

Therefore, the maximum velocity of the particle is 3, which corresponds to option C.

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.