The position of a toy locomotive moving on a

straight track along the x-axis is given by the
equation
x = t4 โˆ’ 6 t2 + 9 t ,
where x is in meters and t is in seconds.
The net force on the locomotive is equal to
zero when t is equal to

x = t^4 - 6t^2 + 9t

Take the derivative to get the velocity.

v = 4t^2 - 12t + 9

Take the derivative again to get acceleration.

a = 8t - 12

Now, let's shift to force formulas.

F = ma

If a = 0, F = 0. Therefore, the net force on the locomotive is equal to zero at the time when a = 0.

a = 0 = 8t -12
8t -12 = 0
8t = 12
t = 12/8 = 1.5 seconds

Well, to find when the net force on the locomotive is equal to zero, we need to look for the times when the acceleration is zero. In this case, the net force is zero when the second derivative of the position equation is equal to zero. So, let me crunch some numbers here... *beep boop beep*

Alright, after doing the math, it appears that the second derivative of the position equation is:

x''(t) = 12t^2 - 12

Setting this equal to zero, we have:

12t^2 - 12 = 0

Simplifying that, we get:

t^2 - 1 = 0

Now solving for t, we find that when t is equal to +1 or -1, the net force on the locomotive is equal to zero. But we're dealing with toy trains here, so my bet is that the answer is a positive one. So, the net force is equal to zero when t is equal to 1 second.

Now if only we could get that toy locomotive to move without any force... Maybe we could use some clown magic? ๐Ÿคกโœจ

To find the value of t when the net force on the locomotive is equal to zero, we need to find the time at which the acceleration of the locomotive is zero. In other words, we need to find the time when the second derivative of the position equation is zero.

The second derivative of x with respect to t is obtained by differentiating the equation twice:

x(t) = t^4 - 6t^2 + 9t
First derivative: x'(t) = 4t^3 - 12t + 9
Second derivative: x''(t) = 12t^2 - 12

Setting x''(t) equal to zero and solving for t:

12t^2 - 12 = 0
12t^2 = 12
t^2 = 1
t = ยฑ1

So, the net force on the locomotive is equal to zero when t is equal to t = ยฑ1 second.

To find when the net force on the locomotive is equal to zero, we need to determine the value(s) of t that make the equation x = t^4 โˆ’ 6t^2 + 9t equal to zero.

Step 1: Set the equation x = t^4 โˆ’ 6t^2 + 9t equal to zero.
t^4 โˆ’ 6t^2 + 9t = 0

Step 2: Factor the equation (if possible).
Since the equation is a quadratic equation in t^2, we can factor it as follows:
t(t^3 - 6t + 9) = 0

Step 3: Apply the zero product property.
Set each factor equal to zero and solve for t.
t = 0 or t^3 - 6t + 9 = 0

Step 4: Solve the equation t^3 - 6t + 9 = 0.
Unfortunately, there is no simple algebraic method to solve this equation, as it is a cubic equation. However, we can use numerical methods to approximate the solutions.

One common numerical method is the Newton-Raphson method. By repeatedly applying this method, we can estimate the values of t that make the equation equal to zero.

Alternatively, graphing the equation and visually inspecting the graph can provide approximate values of t that make the equation equal to zero.

Using a graphing calculator or graphing software, plot the graph of the equation y = t^3 - 6t + 9. The x-intercepts of the graph, where y = 0, will correspond to the values of t that make the equation equal to zero.

By finding the intersections of the graph with the x-axis or using the appropriate technology, we can obtain the values of t where the net force on the locomotive is equal to zero.

Note: Without the specific constraints of the problem or the use of numerical methods, it may not be possible to find the exact value(s) of t where the net force is zero.