



 
Hi, The key to this problem is that average rate is distance divided by time. Suppose the forward trip takes $t$ hours then for this trip \[x = \frac{720}{t} \mbox{ kilometers per hour.}\] How long does it take for the return trip? Write an equation, similar to the one above for the return trip. This rate (speed) is 10 kilometers per hour larger than the rate for the forward trip. This gives you an equation in $t.$ Solve it for $t.$ Once you know $t$ you can calculate the rates but make sure you convert them to meters per second. Penny  


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