the velocity of a particle moving along the x-axis at any time t>=0 is given by
v(t)=cos(pi t)-t(7-2pi)
find the acceleration at any time t
since a = dv/dt
v = cos(πt) - 7t + 2πt
a = =πsin(πt) - 7 + 2π
shouldn't it be a=-pi sin(pi t)-7+2pi then?
Yes, just noticed that I must have hit the = key instead of the -
notice I have ==
good catch!
To find the acceleration at any time t, we need to take the derivative of the velocity function with respect to time.
Given that the velocity function is v(t) = cos(pi t) - t(7 - 2pi), we can differentiate it with respect to t.
1. To find the derivative of cos(pi t), we use the chain rule. The derivative of cos(pi t) with respect to t is -sin(pi t) times the derivative of pi t with respect to t, which is pi.
So, the derivative of cos(pi t) with respect to t is -pi sin(pi t).
2. To find the derivative of t(7 - 2pi), we apply the product rule. The derivative of t with respect to t is 1, and the derivative of 7 - 2pi with respect to t is 0 (since it's a constant).
Using the product rule, we have (t * 0) + (1 * (7 - 2pi)), which simplifies to 7 - 2pi.
Now, we can add the derivatives of the two terms to find the acceleration function.
acceleration (a) = derivative of v(t) = derivative of [cos(pi t) - t(7 - 2pi)].
Acceleration (a) = -pi * sin(pi t) + 7 - 2pi.
Therefore, the acceleration at any time t is given by the equation a(t) = -pi * sin(pi t) + 7 - 2pi.