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.
My answers:
a)(5)/(1-t)^2
b).6 meters (I KNOW this is wrong!)
c) (10t)/t(1-t)
I just know my answers are wrong if someone could just tell me where I went wrong?
Your derivative, (the velocity), should have been 10 -5(1+t)^2
now substitute t=6 into that.
your acceleration, the second derivative, should be
10/(1+t)^3
oops, forgot the divistion sign
first derivative is 10-5/(1+t)^2
for the velocity derivative; I get 5/49 which is .102 meters? That still doesn't sound right.
For the acceleration are you taking the 2nd derivative... because if so; why still have the 10?
Let me provide a step-by-step solution to the given problem.
a) To derive a formula for the velocity of the car, you need to find the derivative of the distance function, D(t), with respect to time, t.
D(t) = 10t + (5)/(1+t) - 5
To take the derivative, start by differentiating each term separately. The derivative of 10t is simply 10. For the second term, (5)/(1+t), you can use the quotient rule, which states that for a function u/v, the derivative is (u'v - uv')/(v^2). Applying this rule, we get:
d/dt[(5)/(1+t)] = [(0)(1+t) - 5(1)]/(1+t)^2
= -5/(1+t)^2
The derivative of the constant term, -5, is 0. Therefore, the derivative of D(t) is:
dD(t)/dt = 10 - 5/(1+t)^2
So, the formula for the velocity, V(t), is:
V(t) = 10 - 5/(1+t)^2
b) To find how fast the car is moving at 6 hours, substitute t = 6 into the velocity formula:
V(6) = 10 - 5/(1+6)^2
= 10 - 5/49
= (490 - 5)/49
= 485/49
â 9.898 meters per hour
Therefore, the car is moving approximately at a speed of 9.898 meters per hour after 6 hours.
c) To derive a formula for the car's acceleration, you need to take the derivative of the velocity function, V(t), with respect to time, t.
V(t) = 10 - 5/(1+t)^2
Differentiating this function, we get:
dV(t)/dt = d/dt[10 - 5/(1+t)^2]
= 0 - (-5(2)/(1+t)^3)
= 10/(1+t)^3
Therefore, the formula for the acceleration, A(t), is:
A(t) = 10/(1+t)^3
Your initial answers were incorrect, but I hope this explanation helps clarify the correct solutions.