Suppose that a cyclist began a 305 mi ride across a state at the western edge of the​ state, at the same time that a car traveling toward it leaves the eastern end of the state. If the bicycle and car met after 5 hr and the car traveled 34 mph faster than the​ bicycle, find the average rate of each.

Answer 1

Let the average speed of the cyclist be \( x \) mph. Then, the speed of the car will be \( x + 34 \) mph, since it travels 34 mph faster than the bicycle.

The total distance covered by both the cyclist and the car when they meet is equal to the distance of 305 miles. The time taken by both the cyclist and the car until they meet is 5 hours.

Now we can set up an equation based on the distances traveled by both:

1. The distance traveled by the cyclist in 5 hours is: \[ \text{Distance}_{\text{cyclist}} = \text{Speed} \times \text{Time} = x \times 5 = 5x \]

2. The distance traveled by the car in 5 hours is: \[ \text{Distance}_{\text{car}} = \text{Speed} \times \text{Time} = (x + 34) \times 5 = 5(x + 34) \]

The sum of the distances traveled by the cyclist and the car should equal the total distance of 305 miles: \[ 5x + 5(x + 34) = 305 \]

Now we can simplify the equation: \[ 5x + 5x + 170 = 305 \] \[ 10x + 170 = 305 \]

Now, subtract 170 from both sides: \[ 10x = 305 - 170 \] \[ 10x = 135 \]

Next, divide both sides by 10: \[ x = 13.5 \]

So the average speed of the cyclist is \( 13.5 \) mph.

Now let's determine the speed of the car: \[ \text{Speed}_{\text{car}} = x + 34 = 13.5 + 34 = 47.5 \text{ mph} \]

Therefore, the average rates of each are:

• Cyclist: 13.5 mph
• Car: 47.5 mph