A particle is moving along the x- axis so that its velocity at any time t ≥0 is give by v(t) = (2 pi - 5)t - sin(pi t)
A. Find the acceleration at ay 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].
a(t) = dv/dt, so
a(t) = (2pi-5) - pi cos(pi t)
as usual, max/min involve derivatives, so max a(t) occurs where da/dt = 0
da/dt = pi^2 sin(pi t)
which is zero at t = 0,1,2,3
how do we know which is min or max? 2nd derivative of a(t), which is
pi^3 cos(pi t)
so, since cos(pi t) > 0 at t=3/2,
min accel is at t=0,2 and is pi-5
max accel is at t=1,3 and is 3pi-5
To find the acceleration at any time t, we need to differentiate the given velocity function, v(t), with respect to time, t.
So, let's differentiate v(t):
v'(t) = d/dt[(2π - 5)t - sin(πt)]
To differentiate the function, we can use the power rule and the chain rule. The derivative of (2π - 5)t is (2π - 5), and the derivative of sin(πt) is π cos(πt).
Therefore, v'(t) = (2π - 5) - π cos(πt)
So the acceleration at any time t is given by v'(t).
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To find the minimum acceleration of the particle over the interval [0,3], we need to evaluate the acceleration function, v'(t), for all values of t in the given interval and find the minimum value.
Let's plug in the values of t in the interval [0,3] into the acceleration function, v'(t), and find the minimum value.
For t = 0, v'(0) = (2π - 5) - π cos(0) = (2π - 5) - π = 2π - π - 5 = π - 5
For t = 3, v'(3) = (2π - 5) - π cos(3π) = (2π - 5) + π = 3π - 5
To find the minimum value, we compare the values π - 5 and 3π - 5.
Since π - 5 ≈ -0.14 and 3π - 5 ≈ 4.43, π - 5 is the minimum value.
Therefore, the minimum acceleration of the particle over the interval [0,3] is π - 5.
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To find the maximum velocity of the particle over the interval [0,2], we need to evaluate the velocity function, v(t), for all values of t in the given interval and find the maximum value.
Let's plug in the values of t in the interval [0,2] into the velocity function, v(t), and find the maximum value.
For t = 0, v(0) = (2π - 5)(0) - sin(π(0)) = 0 - sin(0) = 0
For t = 2, v(2) = (2π - 5)(2) - sin(π(2)) = (4π - 10) - sin(2π) = 4π - 10
To find the maximum value, we compare the values 0 and 4π - 10.
Since 4π - 10 is always greater than 0 for any value of π, the maximum value is 4π - 10.
Therefore, the maximum velocity of the particle over the interval [0,2] is 4π - 10.
a) a(t) = v'(t) = -2πcos(2πt)
ops that's wrong heres right answer
A. a(t) = d/dt (v(t)) = d/dt [(2π-5)t - sin(πt)] = 2π - 5 - πcos(πt)
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B. To find the minimum a(t), we'll derive it and equal to zero:
d/dt (a(t)) = π²sin(πt) = 0 ---> sinπt = 0 ---> πt = 0 + 2k*π, k belongs to the integers
t = 2kπ/π = 2k
Since the v(t) is for any t>0 and the given interval is from 0 to 3, the only possible value for k is 1, therefore t = 2*1 = 2 s.
Minimum a(t) is a(2) = (2π - 5) - πcos(2π) = 2π - 5 - π = π - 5
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C. Max speed: derive v(t) and equal it to zero.
d/dt (v(t)) = a(t) = 0
2π - 5 - πcos(πt) = 0
πcos(πt) = 2π -5
cos(πt) = 2 - (5/π)
πt = acos (2 - 5/π)
t = (1/π)acos(2 - 5/π) ~ 1.63 s
v(0.366) = (2π - 5)*1.63 - sin(π*1.63) ~ 3.01 m/s
DISREGARD MY FIRST POST