A particle moving along x-axis has velocity v(t) = sin(4t) at time t. If the particle is at x=4 when t=0, determine the position of the particle when t=pi/2.
To find the position of the particle when t = π/2, we need to integrate the velocity function v(t) = sin(4t) with respect to time t to get the position function x(t).
The position function x(t) can be found by taking the integral of the velocity function v(t) with respect to t:
x(t) = ∫[v(t)] dt
Integrating sin(4t) with respect to t, we get:
x(t) = ∫sin(4t) dt
To integrate sin(4t), we can use the following trigonometric identity:
∫sin(mx) dx = -1/m * cos(mx)
Letting u = 4t, we have du = 4 dt.
Substituting back in the original integral, we get:
x(t) = ∫sin(u) * (1/4) du
x(t) = (1/4) * (-cos(u)) + C
where C is the constant of integration.
Now we can substitute back in u = 4t:
x(t) = (1/4) * (-cos(4t)) + C
Given that the particle is at x = 4 when t = 0, we can determine the value of the constant C.
When t = 0, we have:
x(0) = (1/4) * (-cos(0)) + C
4 = (1/4) * (-1) + C
4 = -1/4 + C
C = 4 + 1/4 = 17/4
Therefore, the position function x(t) becomes:
x(t) = (1/4) * (-cos(4t)) + 17/4
To find the position of the particle when t = π/2, we substitute t = π/2 into the expression for x(t):
x(π/2) = (1/4) * (-cos(4(π/2))) + 17/4
x(π/2) = (1/4) * (-cos(2π)) + 17/4
x(π/2) = (1/4) * (-1) + 17/4
x(π/2) = -1/4 + 17/4
x(π/2) = 16/4
x(π/2) = 4
Therefore, the position of the particle when t = π/2 is x = 4 units.
To find the position of the particle when t = π/2, we need to integrate the given velocity function v(t) = sin(4t) with respect to t.
The position function x(t) is given by integrating the velocity function over the time interval [0, t].
x(t) = ∫[0, t] v(t) dt
Integrating sin(4t) with respect to t will give us the position function x(t):
x(t) = ∫[0, t] sin(4t) dt
To solve this integral, we can use a simple u-substitution. Let's let u = 4t.
Then, du = 4 dt.
Rearranging, we have dt = du/4.
Substituting this back into our integral:
x(t) = ∫[0, t] sin(u) (du/4)
x(t) = (1/4) ∫[0, t] sin(u) du
The integral of sin(u) is -cos(u), so we have:
x(t) = (1/4)(-cos(u)) + C
Applying the limits of integration [0, t]:
x(t) = (1/4)(-cos(4t)) + C
We can determine the constant C by using the initial condition x(0) = 4.
x(0) = (1/4)(-cos(4(0))) + C
Since cos(0) = 1, we can simplify this to:
4 = (1/4)(-1) + C
Solving for C:
C = 4 + (1/4) = 17/4
Therefore, the position function for the particle is:
x(t) = (1/4)(-cos(4t)) + 17/4
Now, to find the position of the particle when t = π/2, we substitute t = π/2 in the position function:
x(π/2) = (1/4)(-cos(4(π/2))) + 17/4
Simplifying further:
x(π/2) = (1/4)(-cos(2π)) + 17/4
Since cos(2π) = 1, we have:
x(π/2) = (1/4)(-1) + 17/4
x(π/2) = -1/4 + 17/4
x(π/2) = 16/4 = 4
Therefore, the position of the particle when t = π/2 is 4 units.
ayt
v(t) = ds/dt, so
s(t) = -1/4 cos(4t) + C
s(0) = -1/4 + C = 4, so C = 17/4
s(π/2) = -1/4 cos(π/2) + 17/4 = 17/4