A particle moves along the x-axis and its position for time t is greater than or equal to 0, is s (t)=cos(2t)+sec(t). When t=pi, the acceleration of the particle is
d^2s/dt^2 = -4cos(2t)+sec^3(t)+sec(t)tan^2(t)
now just plug and chug.
To find the acceleration of the particle at a given time, we need to take the second derivative of the position function with respect to time.
Given: s(t) = cos(2t) + sec(t)
First, let's find the velocity function by taking the derivative of the position function with respect to time.
v(t) = d/dt[s(t)] = d/dt[cos(2t) + sec(t)]
To find the derivative of cos(2t), we use the chain rule:
d/dt[cos(2t)] = -2sin(2t)
To find the derivative of sec(t), we use the fact that sec(t) = 1/cos(t):
d/dt[sec(t)] = d/dt[1/cos(t)] = -sec(t) * tan(t)
Therefore:
v(t) = -2sin(2t) + (-sec(t) * tan(t))
Now, let's find the acceleration function by taking the derivative of the velocity function with respect to time.
a(t) = d/dt[v(t)] = d/dt[-2sin(2t) + (-sec(t) * tan(t))]
To find the derivative of -2sin(2t), we use the chain rule:
d/dt[-2sin(2t)] = -4cos(2t)
To find the derivative of -sec(t) * tan(t), we use the product rule:
d/dt[-sec(t) * tan(t)] = -sec(t) * sec(t) - sec(t) * tan^2(t) = -sec^2(t)(1 + tan^2(t))
Therefore:
a(t) = -4cos(2t) - sec^2(t)(1 + tan^2(t))
Now, let's find the acceleration at t = π by substituting π into our acceleration function.
a(π) = -4cos(2π) - sec^2(π)(1 + tan^2(π))
Since cos(2π) = 1 and tan(π) = 0, we can simplify further:
a(π) = -4(1) - sec^2(π)(1 + 0)
a(π) = -4 - sec^2(π)
Finally, we can substitute sec^2(π) = 1 into the equation:
a(π) = -4 - 1
a(π) = -5
Therefore, the acceleration of the particle at t = π is -5.