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 the straight-line motion described by the equation, we need to take the derivative of the equation with respect to time.
1. Start with the equation: s = (t+2)^4 - (t+2)^(-2).
2. Compute the derivative of s with respect to t. To do this, we will use the power rule for differentiation.
ds/dt = 4(t+2)^3 - (-2)(t+2)^(-3).
3. Substitute the given time t = 1 into the derivative expression.
ds/dt = 4(1+2)^3 - (-2)(1+2)^(-3).
4. Simplify the expression:
ds/dt = 4(3)^3 - (-2)(3)^(-3).
ds/dt = 108 - (-2/27).
ds/dt = 108 + 2/27.
ds/dt = 108.074.
The instantaneous velocity at t = 1 is approximately 108.1 units.
5. To find the acceleration, take the derivative of the velocity equation:
d^2s/dt^2 = 12(t+2)^2 - (-6)(t+2)^(-4).
6. Substitute t = 1 into the second derivative expression:
d^2s/dt^2 = 12(1+2)^2 - (-6)(1+2)^(-4).
d^2s/dt^2 = 12(3)^2 - (-6)(3)^(-4).
d^2s/dt^2 = 12(9) - (-6/81).
d^2s/dt^2 = 108 - (-6/81).
d^2s/dt^2 = 108 + 0.074.
d^2s/dt^2 = 108.074.
The instantaneous acceleration at t = 1 is approximately 108.1 units.
Therefore, the instantaneous velocity at t = 1 is approximately 108.1 units, and the instantaneous acceleration at t = 1 is also approximately 108.1 units.