A particle is moving along the x-axis so that at time t its acceleration is
a(t)=ðcos(ðt)
At time t=1/2, the velocity v of the particle is 1/2.
Find the velocity of the particle at any time t. I think I got that right, as sinðt + C.
Now it wants the minimum velocity of the particle. How do I find that?
Ugh the pi symbol didn't work. All of the ð should be pi. so it would be a(t) = picos(pit)
and my answer should be sin(pit) + C
a(t) = π cos(πt)
v(t) = sin (πt) + c . where v(t) is velocity
when t=1/2, v(1/2) = 1/2
1/2 = sin (π/2) + c
1/2 = 1 + c
c = -1/2
v(t) = sin (πt) - 1/2
for a min of v(t) , v'(t) = 0 , that is, a(t) = 0
πcos(πt) = 0
cos(πt) = 0
πt = π/2 or πt = 3π/2
t = 1/2 or t = 3/2
we already know v(1/2) = 1/2
we need
v(3/2) = sin(3π/2) - 1/2
= -1 - 1/2 = -3/2
so the minimum velocity is -3/2
To find the minimum velocity of the particle, we need to find the times when the velocity is at its lowest. This corresponds to the points where the velocity function reaches its minimum value.
First, let's integrate the acceleration function to obtain the velocity function:
v(t) = ∫ a(t) dt = ∫ ðcos(ðt) dt
Using the integral of the cosine function, we get:
v(t) = ðsin(ðt) + C
Now, we are given that the velocity of the particle at t = 1/2 is 1/2. Substituting these values into the velocity function:
1/2 = ðsin(ð/2) + C
To find the constant C, we can solve this equation for C:
C = 1/2 - ðsin(ð/2)
Therefore, the velocity of the particle at any time t is:
v(t) = ðsin(ðt) + (1/2 - ðsin(ð/2)))
To find the minimum velocity, we need to find the lowest value of the velocity function. Since the sine function's range is [-1, 1], the minimum value of ðsin(ðt) occurs when ðsin(ðt) = -1.
So, we have:
-1 = ðsin(ðt)
Dividing both sides by ð:
-1/ð = sin(ðt)
Using inverse trigonometric functions, we find:
ðt = arcsin(-1/ð)
Simplifying, we get:
t = arcsin(-1/ð) / ð
Now, substitute this value of t back into the velocity function:
v(min) = ðsin(ðt) + (1/2 - ðsin(ð/2)))
v(min) = ðsin(arcsin(-1/ð) / ð) + (1/2 - ðsin(ð/2)))
Therefore, the minimum velocity of the particle is ðsin(arcsin(-1/ð) / ð) + (1/2 - ðsin(ð/2))).
To find the minimum velocity of the particle, you need to find the critical points of the velocity function.
First, let's integrate the acceleration function a(t) to find the velocity function v(t). Given that a(t) = πcos(πt), integrating once will give you:
∫a(t) dt = ∫πcos(πt) dt
Integrating cos(πt) with respect to t results in sin(πt), so the integral becomes:
v(t) = sin(πt) + C
Now, we know that at t = 1/2, the velocity v of the particle is 1/2. Plugging this value into the velocity function:
1/2 = sin(π/2) + C
Since sin(π/2) equals 1, we can solve for C:
1/2 = 1 + C
C = -1/2
Now we have the velocity function v(t) = sin(πt) - 1/2.
To find the minimum velocity, we need to locate the critical points of the velocity function. A critical point occurs where the derivative equals zero or does not exist.
Taking the derivative of v(t):
v'(t) = πcos(πt)
Setting v'(t) equal to zero:
πcos(πt) = 0
cos(πt) = 0
The cosine function is equal to zero at t = 1/2, t = 3/2, t = 5/2, and so on. These values correspond to the local minimum points of the velocity function.
To find the minimum velocity, we can substitute these values of t back into the velocity function v(t):
v(1/2) = sin(π/2) - 1/2 = 1 - 1/2 = 1/2 - 1/2 = 0
Therefore, the minimum velocity of the particle is 0.