The equation of motion of a particle is
s = t3 − 12t, where s is in meters and t is in seconds.
For the velocity function, I got v(t)=3t^2-12 and for acceleration, a(t)=6t. The final question is: Find the acceleration when the velocity is 0. How do I do that?
well, let's set the velocity equal to zero ...
3t^2 - 12 = 0
3t^2 = 12
t^2 = 4
t = ± 2, but t ≥0
so the velocity is zero when t = 2
so what is the acceleration when t = 2 ??
s=t^3-12t
v=ds/dt=3t^2-12
a=dv/dt=6t
when is velocity 0?
v=3t^2-12=0
t^2=4
t=2 and at t=2, a=6*2=12
To find the acceleration when the velocity is 0, we need to determine the value of t when v(t) equals 0.
Given that v(t) = 3t^2 - 12, we can set it equal to 0 and solve for t:
3t^2 - 12 = 0
Now, let's solve this equation:
3t^2 = 12
Divide both sides by 3:
t^2 = 4
Taking the square root of both sides:
t = ±2
So, we have two possible values for t: t = 2 and t = -2.
However, since time (t) cannot be negative in this context, we can disregard the solution t = -2.
Thus, the acceleration when the velocity is 0 is given by plugging the value t = 2 into the acceleration function a(t):
a(t) = 6t
a(2) = 6(2) = 12
Therefore, when velocity is 0, the acceleration is 12 m/s^2.
To find the acceleration when the velocity is 0, you need to set the velocity function equal to zero and solve for the corresponding value of t. Then, substitute the value of t into the acceleration function to find the acceleration at that time.
Step 1: Set the velocity function equal to zero:
v(t) = 3t^2 - 12 = 0
Step 2: Solve for t by factoring or using the quadratic formula:
3t^2 - 12 = 0
t^2 - 4 = 0
(t - 2)(t + 2) = 0
This equation has two solutions: t = 2 and t = -2.
Step 3: Substitute the value of t into the acceleration function:
a(t) = 6t
For t = 2:
a(2) = 6(2) = 12 m/s^2
For t = -2:
a(-2) = 6(-2) = -12 m/s^2
Therefore, the acceleration when the velocity is 0 can be either +12 m/s^2 or -12 m/s^2, depending on the direction of motion.