After t hours of an 8-hour trip the distance a car travels is modeled by:

D(t)= 10t + (5)/(1+t) - 5
where D(t) is measured in meters.

a) derive a formula for the velocity of the car.
b) how fast is the car moving at 6 hours?
c) derive a formula for the car's acceleration.

I'm sorry bobpursley-I just wanted to ask you about part b; I got .102 m. Is this wrong or right; because it sounds wrong. I plugged 6 into the first derivative of the original function and then found part c with the 2 derivative. Is this correct?
Thank You.

Of course it is wrong, your part a (the velocity) was wrong.

I hope you used the correct first derivative to get the second.

To derive a formula for the velocity of the car, we need to find the derivative of the distance function, D(t), with respect to time, t.

a) To find the velocity, we need to find the derivative of D(t). So, let's differentiate D(t) with respect to t:

D'(t) = d/dt (10t + (5)/(1+t) - 5)

Differentiating each term separately:
The derivative of 10t with respect to t is simply 10.
The derivative of (5)/(1+t) with respect to t involves using the quotient rule. The quotient rule states that if you have a function in the form f(t)/g(t), the derivative is given by [(g(t) * f'(t)) - (f(t) * g'(t))] / [g(t)]^2. In this case, f(t) = 5 and g(t) = (1+t). So, applying the quotient rule, we get:

d/dt [(5)/(1+t)] = [((1+t) * 0) - (5 * 1)] / [(1+t)^2]
= -5 / (1+t)^2

The derivative of -5 with respect to t is simply 0.

Therefore, combining the derivatives, we have:

D'(t) = 10 - 5 / (1+t)^2

b) To find how fast the car is moving at 6 hours, we need to evaluate the velocity at t = 6. This means plugging in t = 6 into the derivative D'(t):

D'(6) = 10 - 5 / (1+6)^2
= 10 - 5 / 49
= 10 - 5/49
= 490/49 - 5/49
= 485/49
≈ 9.90 m/hr

Therefore, the car is moving at approximately 9.90 meters per hour at 6 hours.

c) To derive a formula for the car's acceleration, we need to find the derivative of the velocity function, D'(t), with respect to time, t.

Taking the derivative of D'(t) with respect to t, we get:

D''(t) = d/dt [10 - 5 / (1+t)^2]

The derivative of 10 with respect to t is 0.

The derivative of -5 / (1+t)^2 involves using the chain rule. The chain rule states that if you have a function g(f(t)), the derivative with respect to t is given by g'(f(t)) * f'(t). In this case, g(u) = -5/u^2 and f(t) = (1+t). So, using the chain rule, we have:

d/dt [-5 / (1+t)^2] = (-5) * d/dt [(1+t)^(-2)]
= (-5) * (-2) * (1+t)^(-2-1)
= 10 / (1+t)^3

Therefore, the formula for the car's acceleration, D''(t), is:

D''(t) = 10 / (1+t)^3