Paul ride his bike to school and take him 30 mins at the speed of 20kph Paul go home from school at 4:00pm half Way during his travel his bike broke and decide to walk home Paul reached at exactly 5:00pm calculate the walking speed of Paul?
Answer (2)
Answer:The slope-intercept form of the equation is: = −1/2 − 4Slope (m): −1/2Y-intercept (b): −4Step-by-step explanation:x+2y=−8Step 1: Solve for Subtract from both sides:2 = − − 8Divide by 2: = − 1/2 − 4
[tex]\text{Given:} \\[/tex][tex]\text{Time to school (bike)} = 30 \text{ minutes} = 0.5 \text{ hours} \\[/tex][tex]\text{Speed to school (bike)} = 20 \text{ kph} \\[/tex][tex]\text{Arrival time at home} = 5:00 \text{ pm} \\[/tex][tex]\text{Find:} \\[/tex][tex]\text{Walking speed} \\[/tex][tex]\text{Solution:} \\[/tex][tex]\text{Distance to school} = \text{Speed} \times \text{Time} = 20 \text{ kph} \times 0.5 \text{ hours} = 10 \text{ km} \\[/tex][tex]\text{Distance to halfway point} = \frac{10 \text{ km}}{2} = 5 \text{ km} \\[/tex][tex]\text{Time to halfway point} = \frac{5 \text{ km}}{20 \text{ kph}} = 0.25 \text{ hours} = 15 \text{ minutes} \\[/tex][tex]\text{Time Paul started walking} = 4:00 \text{ pm} + 15 \text{ minutes} = 4:15 \text{ pm} \\[/tex][tex]\text{Time taken to walk} = 5:00 \text{ pm} - 4:15 \text{ pm} = 45 \text{ minutes} = 0.75 \text{ hours} \\[/tex][tex]\text{Walking speed} = \frac{5 \text{ km}}{0.75 \text{ hours}} \approx 6.67 \text{ kph} \\[/tex][tex]\text{Answer:} \\[/tex][tex]\text{Paul's walking speed is approximately } 6.67 \text{ kph}.[/tex]
