To compare the speeds of the three cars after 5 hours, let's first analyze the equations and the information given for each car.
Car 1:
- The distance is represented by the equation \( y = t \).
- This indicates that the car travels 1 mile for every hour. Thus, after 5 hours, the distance traveled would be \( y = 5 ; \text{miles} \).
- The speed of Car 1 is \( 1 ; \text{mile/hour} \).
Car 2:
- The values given suggest a quadratic relationship for distance. Each \( t \) value corresponds to \( t^2 \):
- For \( t = 0 \), \( y = 0 \)
- For \( t = 0.2 \), \( y = 0.04 \)
- For \( t = 0.4 \), \( y = 0.16 \)
- For \( t = 0.6 \), \( y = 0.36 \)
- For \( t = 0.8 \), \( y = 0.64 \)
- For \( t = 1 \), \( y = 1 \)
This suggests the distance equation to be \( y = t^2 \).
- After 5 hours, the distance traveled would be \( y = 5^2 = 25 ; \text{miles} \).
- The speed of Car 2, based on its distance function, is not constant; it varies as speed increases with time. After 5 hours, its speed is \( 10 ; \text{miles/hour} \) since speed can be calculated as the derivative \( \frac{dy}{dt} = 2t \) at \( t = 5 \).
Car 3:
- The distance is represented by the equation \( y = 2t - 1 \).
- After 5 hours, the distance traveled would be \( y = 2(5) - 1 = 10 - 1 = 9 ; \text{miles} \).
- The speed of Car 3 is constant at \( 2 ; \text{miles/hour} \) because it is a linear function.
Comparison After 5 Hours:
- Car 1: 5 miles at 1 mile/hour.
- Car 2: 25 miles at 10 miles/hour (increasing speed).
- Car 3: 9 miles at 2 miles/hour.
In conclusion, Car 2 is the fastest, traveling 25 miles in 5 hours, followed by Car 3 with 9 miles, and then Car 1 with only 5 miles.