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Question

James can jog twice as fast as he can walk. He jogged the first 9 miles to his aunts house, but then he tried and walked the remaining 1 miles. if the total trip took 1.1 hours then what was his average jogging speed?

Answer 1

Check this out.

Let denote the speed of walking V. Then, the speed of jogging is 2V.

We know that the person jogged 9 miles with speed 2V. Then, the time for jogging is 9/(2V).

The same for walking: 1 mile walking and the speed is V. Time for walking is 1/V.

Time for jogging + time for walking = 1.1 hours.

9/(2V) + 1/V = 1.1

When solving for V, we find the speed of walking is V = 5 mph. The speed for jogging is 10 mph.

Answer 2

In algebra, we often put an unknown as x. We don't know what the speed for walking or jogging is but we know that he jogs twice as fast as walking. Let walking be x.

x + 2x = 66 minutes. 3x = 66 or x is 22. 66 - 22 = 44 which is the time he spent jogging. 44 / 9 will be the average speed for the 9 miles that he jogged.

Answer 3

Average speed = (total distance) / (total time)

regardless of how you traveled and what actual velocities you used.

Average speed = 10 miles / 1.1 hours = 9.09 miles/hour