Posted by **George** on Saturday, January 10, 2009 at 6:30pm.

After 35 minutes of jogging, at the 9-km point in a total 10-km race, jogger 1 is behind the leader and moving at the same speed. What would have to be your accelearation in order to catch up to the finish line? Assuming the leader maintains a constant speed the rest of the race.

Is the answer 3.7

- Physics -
**drwls**, Saturday, January 10, 2009 at 6:57pm
1 what behind the leader? 1 km? I will assume so. After 35 minutes, runner 1 would then have 1 km to go and runner 2 would have 2 km to go. The race will be over in 1 km/(9 km/35 min) = 35/9 = 3.89 minutes. Runner 2 must have an average speed over the last 2 km of

2km/3.59 min = 0.557 km/min to catch up. He started the "finishing kick" at a speed of 8 km/35 min = 0.229 km/min and will have to end it at 0.656 km/min. Such a speed is not humanly possible.

I will leave you to compute the required acceleration. (A 0.427 km/min increase in speed in 3.89 minutes)

## Answer this Question

## Related Questions

- Physics - Sorry for rewording it wrong. After 35 minutes of jogging, at the 9-km...
- Physics - After 35 minutes of jogging, at the 9-km point in a total 10-km race, ...
- Physics - After 35 minutes of jogging, at the 9-km point in a total 10-km race, ...
- physics - After 35 minutes of running, at the 8-{\rm km} point in a 9-{\rm km} ...
- physics - After 35 minutes of running, at the 8-{\rm km} point in a 9-{\rm km} ...
- Physics - "A jogger runs with a velocity of 6.0 km/h [25° N of W] for 35 min and...
- Physics - A jogger starting morning run accelerates from a stand still to their ...
- physics - need help with this physics questions a jogger runs with a velocity of...
- Algebra 1 - To train for a race, Will begins by jogging 11 minutes one day per ...
- physics - need help with doing this combining vector question A jogger runs with...

More Related Questions