A particle moves along the x-axis so that at any time t, measured in seconds, its position is given by s(t) = 4sin(t) - cos(2t), measured in feet. What is the acceleration of the particle at time t = 0 seconds?
s" = -4sin(t) + 4cos(2t)
maybe you can at least supply the units, eh?
I think it is -1
I got -1
huh. I guess you don't like my formula. What did you use?
To find the acceleration of the particle at time t = 0 seconds, we need to find the second derivative of the position function, s(t).
The position function is given by s(t) = 4sin(t) - cos(2t).
To find the acceleration, we need to take the derivative twice.
First, let's find the first derivative of s(t):
s'(t) = d/dt (4sin(t) - cos(2t))
We can find the derivative of each term separately, using the derivative rules.
The derivative of sin(t) is cos(t). So, the derivative of 4sin(t) is 4cos(t).
The derivative of cos(2t) is -2sin(2t). So, the derivative of -cos(2t) is 2sin(2t).
Therefore, the first derivative of s(t) is:
s'(t) = 4cos(t) + 2sin(2t)
Now, let's find the second derivative of s(t):
s''(t) = d/dt (4cos(t) + 2sin(2t))
Again, we can find the derivative of each term separately.
The derivative of cos(t) is -sin(t). So, the derivative of 4cos(t) is -4sin(t).
The derivative of sin(2t) is 2cos(2t). So, the derivative of 2sin(2t) is 4cos(2t).
Therefore, the second derivative of s(t) is:
s''(t) = -4sin(t) + 4cos(2t)
Now, let's substitute t = 0 into s''(t) to find the acceleration at t = 0 seconds:
s''(0) = -4sin(0) + 4cos(2(0))
Since sin(0) = 0 and cos(0) = 1, we have:
s''(0) = -4(0) + 4(1) = 0 + 4 = 4
Therefore, the acceleration of the particle at time t = 0 seconds is 4 feet per second squared.