after t seconds the position of a particle which is moving along a straight line is x=2t^3-9t^2+12t+6, when is the acceleration zero? determine the velocity at that time.
v = dx/dt = 6 t^2 - 18 t + 12
a = dv/dt = d^2x/dt^2 = 12 t - 18
a is 0 when t = 18/12 = 3/2 = 1.5 seconds
then v = 6(2.25) - 18(1.5) + 12
To find the time at which the acceleration is zero, we need to find the derivative of the position function with respect to time, and then solve for when this derivative equals zero.
1. Find the derivative of the position function:
The derivative of x with respect to t (denoted as dx/dt or x') will give us the velocity function.
dx/dt = 6t^2 - 18t + 12
2. Next, find the derivative of the velocity function (x') to get the acceleration function (a):
The derivative of x' with respect to t will give us the acceleration.
d²x/dt² = 12t - 18
3. Set the acceleration function (a) to zero and solve for t:
12t - 18 = 0
12t = 18
t = 18/12
t = 1.5 seconds
Therefore, the acceleration is zero at t = 1.5 seconds.
4. Find the velocity at that time (t = 1.5 seconds):
Substitute t = 1.5 into the velocity function:
dx/dt = 6(1.5)^2 - 18(1.5) + 12
= 13.5 - 27 + 12
= -1.5
Therefore, the velocity at t = 1.5 seconds is -1.5 units/s.