The position of a particle is given by the function x = ( 6 t^3 - 3 t^2 + 2) m, where t is in s.
(a) At what time or times is vx = 0 m/s? (Enter NONE for t2 if there is only one time.)
Smaller time, t1 =
Larger time, t2 =
(b) What are the particle's position and its acceleration at this time(s)? position at time t1 =
acceleration at time t1 =
position at time t2 =
acceleration at time t2 =
vx = dx/dt = 18 t^2 -6 t =6t(3t-1)
that is 0 at
t = 0 and at t = 1/3
calculate x at t = 0 and at t = 1/3
then acceleration = 36 t - 6 = 6(6t-1)
calculate acceleration at whatever time
To find the time or times when the velocity of the particle, vx, is equal to zero, we first need to find the expression for vx.
The expression for velocity, v, can be obtained by differentiating the function for position, x, with respect to time, t.
v = dx/dt
Since x = (6t^3 - 3t^2 + 2), let's differentiate it to find v.
v = d(6t^3 - 3t^2 + 2)/dt
Taking the derivative term by term, we get:
v = 18t^2 - 6t
Now, to find the time or times when vx = 0 m/s, we set v equal to zero and solve for t:
0 = 18t^2 - 6t
We can factor out a common factor of 6t:
0 = 6t(3t - 1)
Now, set each factor equal to zero:
6t = 0 or 3t - 1 = 0
From the first equation, we find t = 0.
From the second equation, we find t = 1/3.
So, there are two times when vx = 0 m/s: t1 = 0 and t2 = 1/3.
Now let's find the positions and accelerations at these times.
(a) At time t1 = 0:
To find the position at time t1, we substitute t = 0 into the expression for x:
position at time t1 = x(0) = 6(0)^3 - 3(0)^2 + 2 = 2 m
To find the acceleration at time t1, we can differentiate v with respect to t:
a = dv/dt
a = d(18t^2 - 6t)/dt
Taking the derivative term by term, we get:
a = 36t - 6
Now, substituting t = 0 into a, we find:
acceleration at time t1 = a(0) = 36(0) - 6 = -6 m/s^2
(b) At time t2 = 1/3:
To find the position at time t2, we substitute t = 1/3 into the expression for x:
position at time t2 = x(1/3) = 6(1/3)^3 - 3(1/3)^2 + 2
Simplifying this expression, we get:
position at time t2 = 2 m
To find the acceleration at time t2, we can differentiate v with respect to t:
a = dv/dt
a = d(18t^2 - 6t)/dt
Taking the derivative term by term, we get:
a = 36t - 6
Now, substituting t = 1/3 into a, we find:
acceleration at time t2 = a(1/3) = 36(1/3) - 6 = 0 m/s^2
So, the answers to the questions are:
(a) Smaller time, t1 = 0 s
Larger time, t2 = 1/3 s
(b) Position at time t1 = 2 m
Acceleration at time t1 = -6 m/s^2
Position at time t2 = 2 m
Acceleration at time t2 = 0 m/s^2