the velocity of particle moving along the x axis is given for t>0 by vx=(32t-2t3) m/s, where t is in s. what is the acceleration of the particles when (after t=0) it achieves its maximum displacement in the positive x direction?
max displacement is when v=0, so find t when that is true
a = dv/dt = 32 - 6t^2
To find the acceleration of the particle when it achieves its maximum displacement in the positive x direction, we need to find the time at which the velocity is maximum and then calculate the acceleration at that time.
First, we differentiate the given velocity function with respect to time to find the acceleration:
a = d(vx)/dt
Given vx = 32t - 2t^3, we can differentiate this expression with respect to t:
a = d(32t - 2t^3)/dt
To differentiate, we use the power rule where the derivative of t^n is n*t^(n-1):
a = 32 - 6t^2
Next, to find the time when the velocity is maximum, we set the acceleration equal to zero and solve for t:
32 - 6t^2 = 0
Rearranging the equation gives:
6t^2 = 32
Dividing both sides by 6:
t^2 = 32/6
t^2 = 16/3
Taking the square root of both sides:
t = sqrt(16/3)
t = (4/√3)
Now, to find the acceleration at this time:
a = 32 - 6t^2
Substituting the value of t:
a = 32 - 6*(4/√3)^2
Simplifying further:
a = 32 - 6*(16/3)
a = 32 - (96/3)
a = 32 - 32
a = 0 m/s^2
Therefore, the acceleration of the particle when it achieves its maximum displacement in the positive x direction is zero.
To find the acceleration of the particle when it achieves its maximum displacement in the positive x direction, we need to determine the point at which the particle's velocity is maximum and then calculate the corresponding acceleration.
Given: vx = (32t - 2t^3) m/s
The velocity of the particle is given by the derivative of its displacement with respect to time (i.e., the rate of change of displacement), and the acceleration is given by the derivative of velocity with respect to time.
Step 1: Find the velocity function's maximum point
To find the maximum point of the velocity function, we need to determine where the derivative of the velocity function equals zero.
So, let's differentiate vx with respect to t:
vx = 32t - 2t^3
Differentiating both sides with respect to t:
dvx/dt = 32 - 6t^2
Step 2: Solve for t when dvx/dt = 0
Now, set dvx/dt = 0:
32 - 6t^2 = 0
Rearranging the equation:
6t^2 = 32
Dividing by 6:
t^2 = 32/6
t^2 = 16/3
Taking the square root of both sides:
t = ±√(16/3)
Since t is a positive value for t > 0:
t = √(16/3)
Step 3: Calculate the acceleration at t = √(16/3)
To find the corresponding acceleration, we differentiate the velocity function vx with respect to time t again:
d²vx/dt² = d/dt (32 - 6t^2)
Differentiating the expression:
d²vx/dt² = -12t
Now, substitute t = √(16/3) into the expression:
d²vx/dt² = -12 * √(16/3)
Simplifying:
d²vx/dt² = -12 * (√16/√3)
d²vx/dt² = -12 * (4/√3)
d²vx/dt² = -48/√3
Therefore, the acceleration of the particle when it achieves its maximum displacement in the positive x direction is -48/√3 m/s².