a particle moves along the x-axis so that its velocity at any time t is equal to or > 0 is given by v(t)= (2pie-5)t-sin(piet)
a.find the acceleration at any time t.
b.find the minimum acceleration of the particle over the interval [0,3]
c.find the maximum velocity of the particle over the interval [0,2]
v = (2 pi - 5) t - sin (pi t)
a)
a = dv/dt = (2 pi -5) - pi cos (pi t)
b)
when t = 0, -pi cos (pi t) is -pi and that is as low as a gets
or more formally
da/dt = 0 = pi sin (pi t)
sin (pi t) = 0 when t = 0 , 1 , 3
then
min a = (2 pi -5) - pi = pi - 5
c)
v is max or min when a = 0
(2 pi -5) - pi cos (pi t) = 0
cos (pi t) = 2 - 5/pi = .4084
pi t = 1.15 radians
t = .366
but pi t can also be -1.15 radians which is + 5.13 radians
t = 1.63 than will give me the bigger v
v = (2 pi - 5) t - sin (pi t)
v = (2 pi -5)(1.63) - sin 1.63 pi
= 3
To find the acceleration at any time t, we need to differentiate the velocity function v(t) with respect to time.
a. Differentiating v(t) = (2πe-5)t - sin(πet) with respect to t, we get:
a(t) = d/dt((2πe-5)t - sin(πet))
= 2πe-5 - πe^t * cos(πet)
So, the acceleration at any time t is given by a(t) = 2πe-5 - πe^t * cos(πet).
b. To find the minimum acceleration over the interval [0, 3], we need to find critical points by setting the derivative of acceleration equal to zero:
2πe-5 - πe^t * cos(πet) = 0
Simplifying the equation:
e^t * cos(πet) = 2e-5
Since cos(πet) cannot be negative, we ignore the negative solution. Taking the natural logarithm of both sides:
t = ln(2e-5)
Now, substitute t = ln(2e-5) back into the acceleration function, a(t), to find the minimum acceleration.
c. To find the maximum velocity over the interval [0, 2], we need to find critical points by setting the derivative of the velocity function equal to zero:
v'(t) = 2πe-5 - πe^t * cos(πet) = 0
Solving for t:
πe^t * cos(πet) = 2πe-5
Again, cos(πet) cannot be negative, so we ignore the negative solution. Taking the natural logarithm of both sides:
t = ln(2e-5)
Now, substitute t = ln(2e-5) back into the velocity function, v(t), to find the maximum velocity.
To find the acceleration at any time t, we need to take the derivative of the velocity function with respect to time. The derivative of v(t) gives us the acceleration function, denoted as a(t).
a. To find the acceleration at any time t, take the derivative of the velocity function, v(t):
v(t) = (2πe - 5)t - sin(πet)
Taking the derivative of v(t) with respect to t gives:
a(t) = d/dt[(2πe - 5)t - sin(πet)]
To calculate the derivative, we can apply the power rule for polynomial terms and the chain rule for the sine function.
The derivative of (2πe - 5)t is (2πe - 5).
The derivative of sin(πet) with respect to t is πe * cos(πet) by the chain rule.
Putting it all together, the acceleration function becomes:
a(t) = (2πe - 5) - πe * cos(πet)
b. To find the minimum acceleration of the particle over the interval [0, 3], we need to find the critical points where the acceleration function equals zero, and then determine which one corresponds to the minimum.
Set the acceleration function, a(t), to zero and solve for t:
(2πe - 5) - πe * cos(πet) = 0
Next, solve this equation numerically or graphically to find the value(s) of t that make the acceleration zero.
Once you find the critical point(s), evaluate the acceleration function at each critical point to determine which one gives the minimum acceleration value for the given interval.
c. To find the maximum velocity of the particle over the interval [0, 2], we need to find the critical points where the velocity function reaches its maximum value within the given interval.
To find the critical points, set the derivative of the velocity function, v'(t), equal to zero and solve for t:
v'(t) = (2πe - 5) - πe * cos(πet) = 0
Solve this equation numerically or graphically to find the value(s) of t that produce critical points.
Once you find the critical point(s), substitute these values into the velocity function, v(t), and evaluate it to find the maximum velocity within the given interval.