Final answer:
The velocity and acceleration of a mass attached to a spring with the position function s(t)=5sin(2t) are found using derivatives. The velocity at t=1 is v(1)=10cos(2) inches/second, and the acceleration at t=1 is a(1)=-20sin(2) inches/second².
Step-by-step explanation:
To find the velocity and acceleration of the mass at time t=1 second when it is attached to a vertical spring with a position function s(t)=5sin(2t), we must first take the derivative of the position function to get the velocity, and then take the derivative of the velocity to get the acceleration.
The velocity function is the first derivative of the position function with respect to time, given by v(t) = ds/dt. Taking the derivative of s(t)=5sin(2t) with respect to t gives v(t) = 10cos(2t). To find the velocity at t=1, we substitute 1 into our velocity function: v(1) = 10cos(2) inches/second.
The acceleration function is the second derivative of the position function or the first derivative of the velocity function with respect to time. Calculating the derivative of v(t) = 10cos(2t) gives a(t) = -20sin(2t). Hence, the acceleration at t=1 is a(1) = -20sin(2) inches/second².