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: \( y = t^2 \)
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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 \( t = 1 \), \( y = (1)^2 = 1 \)
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Continuing this pattern will show that after 5 hours: \[ y = 5^2 = 25 \]
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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: \( y = 2t - 1 \)
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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.
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At \( t = 0 \): \( y = 2(0) - 1 = -1 \)
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At \( t = 1 \): \( y = 2(1) - 1 = 1 \)
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At \( t = 2 \): \( y = 2(2) - 1 = 3 \)
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At \( t = 3 \): \( y = 2(3) - 1 = 5 \)
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At \( t = 4 \): \( y = 2(4) - 1 = 7 \)
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At \( t = 5 \): \( y = 2(5) - 1 = 9 \)
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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.