Question

Three cars start traveling at the same time. The distance traveled by each is represented in three different ways, with y representing miles traveled and t representing time passed in hours.
Car 1:
y=t

Car 2: T 0, 0.2, 0.4, 0.6, 0.8, 1
y=t^2 0, 0.04, 0.16, 36.0, 64.0, 1

Car 3:the numbers on the graph show 1 to 5 on the X and Y axis,a curving line on the graph starts from zero curving infinitely upwards on the grapgh near the number 5, this curve is y=2t - 1

Compare the speeds of the 3 cars after 5 hours. In your answer, describe the function represented by each car.

Answer 1

To compare the speeds of the three cars after 5 hours, we first need to analyze the functions representing the distance traveled by each car.

### Car 1:
**Function:** \( y = t \)

- **Description:** This function represents a linear relationship between distance \( y \) and time \( t \). The car travels 1 mile for every hour that passes.
- **Speed:** The speed of Car 1 is constant at 1 mile per hour (mph).

### Car 2:
**Function:** \( y = t^2 \)

- **Description:** This function represents a quadratic relationship. The distance traveled increases with the square of the time. For example:
- At \( t = 0.2 \), \( y = (0.2)^2 = 0.04 \)
- At \( t = 0.4 \), \( y = (0.4)^2 = 0.16 \)
- At \( t = 0.6 \), \( y = (0.6)^2 = 0.36 \)
- At \( t = 0.8 \), \( y = (0.8)^2 = 0.64 \)
- At \( t = 1 \), \( y = (1)^2 = 1 \)

- Continuing this pattern will show that after 5 hours:
\[
y = 5^2 = 25
\]
- **Speed:** The speed of Car 2 increases over time because the distance increases as the square of time. The instantaneous speed (or derivative) can be found to be \( 2t \). After 5 hours, the speed is:
\[
\text{Speed} = 2 \cdot 5 = 10 \text{ mph}
\]

### Car 3:
**Function:** \( y = 2t - 1 \)

- **Description:** This function is linear as well but starts at -1 for \( t = 0 \). The car travels at a rate of 2 miles for every hour that passes.
- At \( t = 0 \): \( y = 2(0) - 1 = -1 \)
- At \( t = 1 \): \( y = 2(1) - 1 = 1 \)
- At \( t = 2 \): \( y = 2(2) - 1 = 3 \)
- At \( t = 3 \): \( y = 2(3) - 1 = 5 \)
- At \( t = 4 \): \( y = 2(4) - 1 = 7 \)
- At \( t = 5 \): \( y = 2(5) - 1 = 9 \)

- **Speed:** The speed of Car 3 is constant at 2 mph.

### Summary:
- **Car 1** travels at **1 mph** (linear).
- **Car 2** travels at an **increasing speed** (quadratic), up to **10 mph** after 5 hours.
- **Car 3** travels at **2 mph** (linear).

Thus, after 5 hours, the speeds are:
- Car 1: 1 mph
- Car 2: 10 mph
- Car 3: 2 mph

Car 2 is the fastest, followed by Car 3, and then Car 1.