Find the instantaneous velocity and acceleration at the given time for the straight-line motion described by the equation: s= (t+2)^4- (t+2)^(-2) at t=1

Round to one decimal place.

To find the instantaneous velocity and acceleration at a given time for a given equation of motion, we need to find the derivative of the equation.

Given the equation: s = (t+2)^4 - (t+2)^(-2)

Step 1: Find the first derivative of the equation with respect to time (t). This will give us the equation for velocity.

Let's find the derivative of the function s(t):
s' = d/dt [ (t+2)^4 - (t+2)^(-2) ]

To find the derivative of each term, we'll use the power rule of differentiation.

The derivative of (t+2)^4 can be found by bringing down the exponent and multiplying it by the term itself:
d/dt [ (t+2)^4 ] = 4(t+2)^(4-1) = 4(t+2)^3

To find the derivative of (t+2)^(-2), we'll use the chain rule. Let u = t+2, then the derivative of u^(-2) with respect to u is: (-2)u^(-3). Now we can multiply it with du/dt (the derivative of t+2), which is simply 1:
d/dt [ (t+2)^(-2) ] = (-2)(t+2)^(-3) * 1 = -2(t+2)^(-3)

Now we can substitute these derivatives back into the equation for s':
s' = 4(t+2)^3 - 2(t+2)^(-3)

Step 2: Evaluate the derivative at the given time, which is t=1.

Substitute t=1 into the equation for s' to find the velocity at t=1:
v(t=1) = 4(1+2)^3 - 2(1+2)^(-3)

Simplifying further, we have:
v(t=1) = 4(3)^3 - 2(3)^(-3)

Now, you can calculate the velocity at t=1 using a calculator or by performing the calculations manually.

Step 3: Find the second derivative of the equation to determine acceleration.

To find the acceleration, we take the derivative of the velocity function (s') with respect to time (t).

a(t) = d^2/dt^2 [ s(t) ]
a(t) = d/dt [ 4(t+2)^3 - 2(t+2)^(-3) ]

Differentiating each term separately:
a(t) = d/dt [ 4(t+2)^3 ] - d/dt [ 2(t+2)^(-3) ]

Using the power rule and chain rule again, we get:

a(t) = 12(t+2)^2 - (-2)(-3)(t+2)^(-4)

Simplifying further:
a(t) = 12(t+2)^2 + 6(t+2)^(-4)

Now, substitute t=1 into the equation for a(t) to find the acceleration at t=1:
a(t=1) = 12(1+2)^2 + 6(1+2)^(-4)

Simplifying further,
a(t=1) = 12(3)^2 + 6(3)^(-4)

Now, you can calculate the acceleration at t=1 using a calculator or by performing the calculations manually. Round the answers to one decimal place if necessary.