A runner goes for a long run in the mountains covering 25 kilometres in five hours. Her distance is described by the function d(t)=2/3t^3-35/6t^2+35/2t where d(t) is in km and t is in hours, 0 less than or equal to t less than or equal to 5 The derivative function is given as d'(t)=2t^2-35/3t+35/2

(a) Find the distance covered after 2.5 hours. (b) Verify that the runner completes 25 km at 5 hours. (c) Find the runner's average speed during the first five hours. (d) Give the units of d' (t) . (e) Find the runner's speed at the start and end of the run. (f) Find the time when the runner is going the fastest and the runner's maximum speed. (g) Find the runner's minimum speed and the time when this occurs.

Given the function \( d(t) = \frac{2}{3}t^3 - \frac{35}{6}t^2 + \frac{35}{2}t \) for the distance covered in kilometers, and its derivative \( d'(t) = 2t^2 - \frac{35}{3}t + \frac{35}{2} \) for the speed, let's solve each part step by step.

### (a) Find the distance covered after 2.5 hours.

We simply substitute \( t = 2.5 \) into the distance function \( d(t) \):

\[
d(2.5) = \frac{2}{3}(2.5)^3 - \frac{35}{6}(2.5)^2 + \frac{35}{2}(2.5)
\]

Let’s calculate this value.

The distance covered after 2.5 hours is approximately 17.71 km.

### (b) Verify that the runner completes 25 km at 5 hours.

We substitute \( t = 5 \) into the distance function \( d(t) \) to check if the runner completes 25 km:

\[
d(5) = \frac{2}{3}(5)^3 - \frac{35}{6}(5)^2 + \frac{35}{2}(5)
\]

Let’s calculate this next.The distance covered after 5 hours is 25 km, confirming that the runner completes the full distance.

### (c) Find the runner's average speed during the first five hours.

The average speed is given by the total distance divided by the total time:

\[
\text{Average speed} = \frac{d(5)}{5}
\]

Let's compute the average speed.The runner's average speed during the first five hours is 5 km/h.

### (d) Give the units of \( d'(t) \).

Since \( d(t) \) represents distance in kilometers and \( t \) represents time in hours, the derivative \( d'(t) \) represents the rate of change of distance with respect to time, which is **speed**. Therefore, the units of \( d'(t) \) are **km/h** (kilometers per hour).

### (e) Find the runner's speed at the start and end of the run.

To find the speed, we evaluate \( d'(t) \) at \( t = 0 \) (start) and \( t = 5 \) (end). The speed is given by:

\[
d'(t) = 2t^2 - \frac{35}{3}t + \frac{35}{2}
\]

Let's compute the speed at these times.It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can manually calculate the speeds using the formula for \( d'(t) = 2t^2 - \frac{35}{3}t + \frac{35}{2} \).

- For \( t = 0 \) (start), plug in \( t = 0 \) into the speed function to find the initial speed.
- For \( t = 5 \) (end), plug in \( t = 5 \) to get the speed at the end.

Would you like help with these calculations? Additionally, here are some related questions you might find helpful:

1. What is the physical significance of \( d'(t) = 0 \)?
2. How can you find the acceleration of the runner using \( d''(t) \)?
3. Can we predict the speed at any given time using \( d'(t) \)?
4. How does the total distance function change if the runner increases her speed by 10%?
5. What happens to the distance function if the run takes place in 4 hours instead?

**Tip:** The derivative of a function gives you the rate of change, which is especially useful for calculating speed in time-related problems.

A runner goes for a long run in the mountains covering 25 kilometres in five hours. Her distance is described by the function d(t)=2/3t^3-35/6t^2+35/2t where d(t) is in km and t is in hours, 0 ≤ t ≤ 5. The derivative function is given as d'(t)=2t^2-35/3t+35/2. (a) Find the distance covered after 2.5 hours. (b) Verify that the runner completes 25 km at 5 hours. (c) Find the runner's average speed during the first five hours. (d) Give the units of d'(t). (e) Find the runner's speed at the start and end of the run. (f) Find the time when the runner is going the fastest and the runner's maximum speed. (g) Find the runner's minimum speed and the time when this occurs.

Solve a runner's distance and speed problem using calculus. Learn how to calculate distance after 2.5 hours, verify 25 km at 5 hours, and find the runner's average speed, speed at the start and end, as well as the maximum and minimum speeds. This problem involves derivatives and optimization, suitable for Grades 10-12 calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation