the velocity of a particle moving along the axis is given for t>0 by v=(32t -2t^3 ) m/s where t is in s what is the acceleration of a particle when (after t=0 ) it achieves its maximum displacement in the positive x direction
you are given velocity.
acceleration is dv/dt.
displacement max will occur when v is zero (it is changing directions). solve for t when v=0. Then use that t in dv/dt=?
To find the acceleration of the particle when it achieves its maximum displacement in the positive x direction, we need to find the derivative of the velocity function with respect to time.
Given:
v = (32t - 2t^3) m/s
To find the acceleration, we differentiate the velocity function with respect to time (t).
a = dv/dt
Differentiating v = (32t - 2t^3) with respect to t, we get:
a = d/dt(32t - 2t^3)
= 32 - 6t^2
So, the acceleration of the particle when it achieves its maximum displacement in the positive x direction is given by:
a = 32 - 6t^2
To find the time (t) at which the particle achieves its maximum displacement, we need to set the velocity function equal to zero:
32t - 2t^3 = 0
Factorizing the equation, we get:
t(32 - 2t^2) = 0
From this equation, we can see that either t = 0 or 32 - 2t^2 = 0.
Solving 32 - 2t^2 = 0, we get:
-2t^2 = -32
t^2 = 16
t = ±4
Since t cannot be negative for this case, t = 4 is the time when the particle achieves its maximum displacement in the positive x direction.
Substituting t = 4 into the acceleration equation:
a = 32 - 6t^2
a = 32 - 6(4^2)
a = 32 - 6(16)
a = 32 - 96
a = -64 m/s^2
Therefore, the acceleration of the particle when it achieves its maximum displacement in the positive x direction is -64 m/s^2.
To find the acceleration of a particle when it achieves its maximum displacement in the positive x-direction, we need to differentiate the velocity function with respect to time. This will give us the expression for acceleration.
Given that the velocity function is v = (32t - 2t^3) m/s, we can differentiate it to find the acceleration:
a = dv/dt
Differentiating each term separately:
dv/dt = d(32t)/dt - d(2t^3)/dt
The derivative of 32t with respect to t is simply 32, as the derivative of a constant multiplied by t is just the constant.
The derivative of 2t^3 with respect to t can be found using the power rule: d(x^n)/dx = n*x^(n-1). In this case, n = 3.
d(2t^3)/dt = 3*2t^(3-1) = 6t^2
Putting it all together:
a = 32 - 6t^2
Now, the question specifies that the particle achieves its maximum displacement in the positive x-direction. This means the particle reaches its highest point and starts moving back towards the origin.
At the highest point, the velocity will be 0, so we can set:
32t - 2t^3 = 0
Factoring out t from the equation:
t(32 - 2t^2) = 0
This gives us two possible solutions:
t = 0 (which we can discard because it's t > 0)
32 - 2t^2 = 0
Solving the second equation:
2t^2 = 32
t^2 = 16
t = ±√16
t = ±4
Since t > 0, we take the positive value of t: t = 4.
Now, substitute t = 4 into the acceleration equation we derived earlier:
a = 32 - 6t^2
a = 32 - 6(4^2)
a = 32 - 6(16)
a = 32 - 96
a = -64 m/s^2
Therefore, when the particle achieves its maximum displacement in the positive x-direction, the acceleration is -64 m/s^2.