A cyclist rides 30 km at an average speed of 9 km per hour. At what rate must the cyclist cover in the next 10 km in order to bring the overall average speed up to 10 km per hour?
+208 Remember, distance=rate*time
Distance Rate Time
First part of trip 30 9 30/9
Last Part of trip 10 x 10/x
Whole trip 40 10 4
(Time for the first part of the trip) + (time for the second part of the trip) = (total time of trip)
\(\frac{30}{9}+\frac{10}{x}=4 \\30x+90=36x \\-6x=-90 \\x=15\)
so the cyclist's rate must be 15 kph in the next 10 km in order to bring the overall speed up to 10 km per hour
JP
+208 Remember, distance=rate*time
Distance Rate Time
First part of trip 30 9 30/9
Last Part of trip 10 x 10/x
Whole trip 40 10 4
(Time for the first part of the trip) + (time for the second part of the trip) = (total time of trip)
\(\frac{30}{9}+\frac{10}{x}=4 \\30x+90=36x \\-6x=-90 \\x=15\)
so the cyclist's rate must be 15 kph in the next 10 km in order to bring the overall speed up to 10 km per hour
JP
