Question

a mass attached to a vertical spring has position function given by ( s(t)=4sin(4t) ) where ( t ) is measured in seconds and ( s ) in inches. find the velocity at time ( t = 5 ). find the acceleration at time ( t = 5 ). question help: message instructor

Explanation:

Step1: Recall the velocity formula

Velocity \(v(t)\) is the derivative of position \(s(t)\). Using the chain rule, if \(s(t)=A\sin(Bt)\), then \(v(t)=A\times B\cos(Bt)\). Here \(A = 4\) and \(B = 4\), so \(v(t)=16\cos(4t)\)

Step2: Calculate velocity at \(t = 5\)

Substitute \(t = 5\) into \(v(t)\): \(v(5)=16\cos(4\times5)=16\cos(20)\)

Step3: Recall the acceleration formula

Acceleration \(a(t)\) is the derivative of velocity \(v(t)\). Since \(v(t)=16\cos(4t)\), then \(a(t)=16\times(- 4)\sin(4t)=-64\sin(4t)\) (using the chain rule for differentiation of \(\cos(u)\) where \(u = 4t\))

Step4: Calculate acceleration at \(t = 5\)

Substitute \(t = 5\) into \(a(t)\): \(a(5)=-64\sin(4\times5)=-64\sin(20)\)

Answer:

Velocity at \(t = 5\): \(16\cos(20)\) inches per second
Acceleration at \(t = 5\): \(-64\sin(20)\) inches per second²