The position of a model train, in feet along a railroad track, is given by
s(t) = 2.5t + 16
after t seconds.
(a) How fast is the train moving?
ft/sec

(b) Where is the train after 4 seconds?
feet along the track

(c) When will the train be 31 feet along the track?
t =
sec

Given the position function of the model train \( s(t) = 2.5t + 16 \), we can break down the parts of the problem:

### (a) How fast is the train moving?
To find the speed (or velocity) of the train, we take the derivative of the position function \( s(t) \) with respect to \( t \). The derivative represents the rate of change of position, which is the velocity.

\[
v(t) = \frac{ds}{dt} = 2.5
\]

So, the train is moving at a constant speed of **2.5 ft/sec**.

### (b) Where is the train after 4 seconds?
We substitute \( t = 4 \) into the position function \( s(t) \) to find the position of the train after 4 seconds:

\[
s(4) = 2.5(4) + 16 = 10 + 16 = 26 \text{ feet}
\]

Thus, the train is **26 feet** along the track after 4 seconds.

### (c) When will the train be 31 feet along the track?
To find when the train is at 31 feet, we set \( s(t) = 31 \) and solve for \( t \):

\[
31 = 2.5t + 16
\]

Subtract 16 from both sides:

\[
15 = 2.5t
\]

Now, divide by 2.5:

\[
t = \frac{15}{2.5} = 6 \text{ seconds}
\]

The train will be **31 feet** along the track after **6 seconds**.

---

Let me know if you need further explanation or have any questions!

Here are 5 additional related questions:
1. What would happen to the speed if the position function were \( s(t) = 3t + 16 \)?
2. How would you find the train's position at any given time \( t \)?
3. What if the initial position (16 feet) was different? How would that affect the results?
4. How would you interpret the result if the position function had a quadratic term, like \( s(t) = 2.5t^2 + 16 \)?
5. How can we represent the train's acceleration if the velocity is constant?

**Tip**: Always pay attention to whether the velocity function changes with time (for acceleration) or remains constant (for uniform motion).

The position of a model train, in feet along a railroad track, is given by s(t) = 2.5t + 16 after t seconds. (a) How fast is the train moving? ft/sec (b) Where is the train after 4 seconds? feet along the track (c) When will the train be 31 feet along the track? t = sec

Learn how to find the speed of a train, its position after 4 seconds, and when it reaches 31 feet using the linear position function s(t) = 2.5t + 16. This problem covers algebraic concepts and basic differentiation. Suitable for students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation