can you please help me thanks

Question

a mass attached to a vertical spring has position function given by s (t)=5 sin(2t)where t measure in second and 8 inches

find the veloctiy at time t=4

find the acceleration at time t=4

Answer 1

I'm not sure what "and 8 inches" refers to, but if s(t) is the position/displacement as a function of time, then the velocity is just its first derivative, and the acceleration is just the second derivative. This is because the derivative is equal to the instantaneous rate of change, which is the relationship between velocity and position, and is the relationship between acceleration and velocity.

s(t) = 5sin(2t)

v(t) = 10cos(2t)

a(t) = -20sin(2t)

v(4) = 10cos(8)

a(4) = -20sin(8)

Answer 2

Take the 1st derivative of the position function to get the velocity function and tale the second derivative (aka the derivative of the derivative) of the position function to get the acceleration function.

To take the first derivative, use the shortcut rule that if y = sin(x) then y' = cos(x) and if y = cos(x) then y' = -sin(x). The only thing different here is that you must use the chain rule and multiply by the derivative of the inside (2t)' = 2:

s(t) = 5sin(2t)

s'(t) = v(t) = 5cos(2t)•2 = 10cos(2t)

s''(t) = v'(t) = a(t) = -10sin(2t)•2 = -20sin(2t)

Now, plug in t = 4 making sure your calculator is in radian mode.

Note: The units for velocity are inches/sec and the units of acceleration are inches/second²