THe equation of motion of a particle is s = t^3 - 3t, where s is in meters and t is in seconds. Find
(a) the velocity and acceleration as
functions of t
(b) the acceleration after 2 s, and
(c) the acceleration when the velocity is 0
Hh
To find the velocity and acceleration as functions of time, we need to take the derivatives of the given equation for position.
(a) Velocity:
The velocity is the derivative of position with respect to time. Taking the derivative of s = t^3 - 3t with respect to t gives us the velocity function:
v(t) = ds/dt
Let's take the derivative of s(t) term by term:
ds/dt = d/dt (t^3 - 3t)
= 3t^2 - 3
Therefore, the velocity function is:
v(t) = 3t^2 - 3
(b) Acceleration at 2 seconds:
To find the acceleration at a specific time (in this case, t = 2 seconds), we can take the derivative of the velocity function with respect to time:
a(t) = dv/dt
Taking the derivative of v(t) = 3t^2 - 3 with respect to t, we get:
dv/dt = d/dt (3t^2 - 3)
= 6t
Substituting t = 2 into the acceleration function:
a(2) = 6(2)
= 12
Therefore, the acceleration at t = 2 seconds is 12 m/s^2.
(c) Acceleration when velocity is 0:
To find the acceleration when the velocity is 0, we need to solve for t when v(t) = 0:
v(t) = 3t^2 - 3 = 0
Setting the velocity function equal to 0, we can solve for t:
3t^2 - 3 = 0
Dividing both sides by 3, we get:
t^2 - 1 = 0
Factoring the equation:
(t - 1)(t + 1) = 0
Setting each factor equal to 0:
t - 1 = 0 or t + 1 = 0
Solving for t:
t = 1 or t = -1
Therefore, the acceleration is 0 when the velocity is 0 at t = 1 second and t = -1 second.