The position of a toy locomotive moving on a
straight track along the x-axis is given by the
equation
x = t
4 − 6 t
2 + 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
Answer in units of s
To find when the net force on the locomotive is equal to zero, we need to determine the time (t) at which the acceleration (second derivative of position) equals zero.
First, let's find the velocity of the locomotive by taking the derivative of the position equation with respect to time:
v = dx/dt = d/dt(t^4 - 6t^2 + 9t)
= 4t^3 - 12t + 9
Next, let's find the acceleration by taking the derivative of velocity with respect to time:
a = dv/dt = d/dt(4t^3 - 12t + 9)
= 12t^2 - 12
Now, we set the acceleration equal to zero and solve for t:
12t^2 - 12 = 0
Divide both sides by 12:
t^2 - 1 = 0
Factor the equation:
(t - 1)(t + 1) = 0
This gives us two possible values for t: t = 1 and t = -1.
However, since time cannot be negative, we discard t = -1.
Therefore, when t = 1 second, the net force on the locomotive is equal to zero.