At 2:00 PM a car's speedometer reads 30 mi/h. At 2:10 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi/h2. Let v(t) be the velocity of the car t hours after 2:00 PM. Then (v(1/6) − v(0))/(1/6 − 0) = ? .
By the Mean Value Theorem, there is a number c such that 0 < c < ? with v'(c) = ? . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 120 mi/h2
0 < c < 1/6
v'(c) is given by the formula at the end of paragraph one.
To show that the acceleration is exactly 120 mi/h^2 between 2:00 PM and 2:10 PM, we can use the mean value theorem for derivatives.
Let v(t) be the velocity of the car t hours after 2:00 PM. We are given that at 2:00 PM (t = 0), the velocity is 30 mi/h, and at 2:10 PM (t = 1/6), the velocity is 50 mi/h.
We want to find the acceleration, which is the derivative of velocity with respect to time, v'(t).
Using the mean value theorem, there exists a number c between 0 and 1/6 such that:
v'(c) = (v(1/6) - v(0))/(1/6 - 0)
Substituting the given velocities:
v'(c) = (50 - 30)/(1/6 - 0)
Simplifying:
v'(c) = 20/(1/6)
Multiplying both the numerator and denominator by 6:
v'(c) = 120
Hence, the acceleration at some time between 2:00 PM and 2:10 PM is exactly 120 mi/h^2.
To solve this problem, we need to use the Mean Value Theorem (MVT) and the concept of velocity and acceleration. Let's break down the steps to find the required values.
1. The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). In this case, the function is the velocity function v(t), and we need to find the value of c where v'(c) = 120 mi/h².
2. We are given that the car's speedometer reads 30 mi/h at 2:00 PM and 50 mi/h at 2:10 PM, which means the velocity function v(t) is continuous on the closed interval [0, 1/6] and differentiable on the open interval (0, 1/6).
3. Now, let's find the velocity function v(t) by using the given information. We can use the formula for average velocity to find the change in velocity over the interval [0, 1/6]:
average velocity = (change in distance) / (change in time)
At 2:00 PM, the car's speedometer reads 30 mi/h, and at 2:10 PM, it reads 50 mi/h. The change in velocity is 50 - 30 = 20 mi/h, and the change in time is 1/6 - 0 = 1/6 hours.
So, the average velocity over this interval is 20 mi/h / (1/6) hours = 120 mi/h.
Therefore, we have v(1/6) - v(0) = 120 mi/h.
4. Now, we can apply the Mean Value Theorem (MVT) to find the value of c where v'(c) = 120 mi/h². Recall that v'(c) represents the acceleration.
By the MVT, we have v'(c) = (v(1/6) - v(0)) / (1/6 - 0).
Substituting the known values, we get v'(c) = 120 mi/h / (1/6) = 720 mi/h².
Therefore, we have shown that at some time between 2:00 and 2:10 PM, the acceleration is exactly 720 mi/h² (not 120 mi/h² as originally stated).
So, the correct answer is:
(v(1/6) − v(0))/(1/6 − 0) = 720 mi/h².