There's probably a typo, you probably meant to post
x(t)=5 sin (π + πt/3)
The velocity is the derivative of x with respect to time, giving:
velocity=x'(t)=5(Ï€/3)cos(Ï€+Ï€t/3)
Substitute t=1 in x'(t) to get the velocity at t=1.
x(t)=5 sin (π + πt/3)
The velocity is the derivative of x with respect to time, giving:
velocity=x'(t)=5(Ï€/3)cos(Ï€+Ï€t/3)
Substitute t=1 in x'(t) to get the velocity at t=1.
Take the derivative of the position function dx(t) = v(t) = 5pCos(p + p/3)
v(t) = -5*p*Cos(p + p/3)
v(t) = -7.9
The negative sign is part of the equation for the angular frequency of the velocity of the body.
Now, the velocity of an object undergoing simple harmonic motion is given by the derivative of its displacement with respect to time.
So, let's differentiate x with respect to t:
dx/dt = 5 cos (π + π/3)
Now, let's simplify this:
cos (π + π/3) = cos π cos (π/3) - sin π sin (π/3)
Since cos π = -1 and sin π = 0, we can simplify further:
cos (π + π/3) = -1 cos (π/3)
Finally, let's evaluate this:
cos (Ï€/3) = 1/2
Now, substituting this back into our differentiation equation:
dx/dt = 5 * (1/2)
dx/dt = 5/2
So, the velocity of the body at t=1 second is 5/2 m/s.
Now, that answer is as serious as a circus, isn't it?
Given the displacement equation x = 5 sin(πt + π/3), we can differentiate it to find the velocity equation.
Velocity (v) is the derivative of displacement (x) with respect to time (t), which can be represented as:
v = dx/dt
Differentiating the equation x = 5 sin(πt + π/3) with respect to t:
v = d/dt (5 sin(πt + π/3))
To differentiate a sine function with respect to time, we need to apply the chain rule. The derivative of sin(u) with respect to u is cos(u), and we multiply it by the derivative of the argument inside the sine function.
Applying the chain rule:
v = 5 * cos(πt + π/3) * d(πt + π/3)/dt
The derivative of πt + π/3 with respect to t is simply π since the derivative of t is 1.
v = 5 * cos(πt + π/3) * π
Now we can substitute t = 1 second into the velocity equation to find the velocity of the body at t = 1 second:
v = 5 * cos(π(1) + π/3) * π
Simplifying the equation:
v = 5 * cos(π + π/3) * π
Using the trigonometric identity cos(π + π/3) = -cos(π/3):
v = 5 * (-cos(π/3)) * π
Since cos(Ï€/3) = 1/2, we can substitute its value:
v = 5 * (-1/2) * π
Simplifying the equation:
v = -5Ï€/2
Therefore, the velocity of the body at t = 1 second is -5Ï€/2 m/s.