What is the acceleration of Steve Prefontaine during his 10-km race?

[fightboy]

In April 1974, Steve Prefontaine completed a 10.0-km race in a time of 27 min, 43.6 s. Suppose “Pre” was at the 9.00-km mark at a time of 25.0 min even. If he accelerates for 60.0 s and maintains the increased speed for the duration of the race, calculate the acceleration that he had. Assume his instantaneous speed at the 9.00-km mark was the same as his overall average speed at that time.
I kept getting very confused on this problem because there were too many unknowns. I don't know his final speed after acceleration, the distance covered during his acceleration, or the distance covered after he was done accelerating. I've tried solving by using a system of equations but i kept getting to many unknown variables which led me too not get any answers. The only thing i pretty much got is that his instantaneous speed at the 9 km mark would be 9000m/1500 s= 6m/s. But i can't use that to solve for acceleration since we don't know the distance traveled. Can anyone help?

 

[olivermsun]

Try drawing a graph of Pre's speed vs. time. I think the problem will be easier to understand once you look at it visually.

 

[fightboy]

Try drawing a graph of Pre's speed vs. time. I think the problem will be easier to understand once you look at it visually.

I did that but since his speed is unknown after the 9 km mark, i can't really draw anything after that point

 

[olivermsun]

Call it something like "$v_\text{later}$." It doesn't have to be plotted exactly to scale – you want to find out what $v_\text{later}$ is, right?

 

[fightboy]

yes since wouldn't that help me find out what the acceleration is? and ok i did that but i still don't get it:(. All i have on my graph after 9 km is an upward line for the first 60 seconds he accelerated, and then a horizontal line to show that he traveled for the duration at constant speed.

 

[olivermsun]

So the vertical axis of your graph has units of speed (km/s) and the horizontal axis has units of time (s). The distance traveled is speed * time, which is exactly the area under the graph.

You don't know exactly where the horizontal line is, but you know the total area (distance traveled) that should be contained under the ramp and a constant line, right?

 

[fightboy]

So the vertical axis of your graph has units of speed (km/s) and the horizontal axis has units of time (s). The distance traveled is speed * time, which is exactly the area under the graph.

You don't know exactly where the horizontal line is, but you know the total area (distance traveled) that should be contained under the ramp and a constant line, right?

Yes the total distance would be 1 km. but I don't know how to divide the distances evenly between the time he accelerated and the time he remained at constant speed till the end of the race.

 

[Simon Bridge]

You know the time he spent accelerating - mark that in on your time axis and draw a dotted line up to the v(t) line.
That should divide the area into two shapes.

You know the equations for the areas of the shapes you can see, you also know how to find the slope of a graph.

 

[fightboy]

Thank you guys very much! The graph really helped once I divided the area in different shapes.

 

[Simon Bridge]

It's a powerful tool - and means you don't have to memorize 5 suvat equations.
Well done.