Simple Acceleration-Distance runner problem doesn't self-check

[Head_Unit]

Hi Folks: I am puzzled by this problem; there must be something wrong but I can't see what...

1. The problem statement: "A runner enters the last 200 m of a race running 6 m/s. She accelerates at 0.2 m/s2 &crosses the finish line at 10 m/s. What is her time for the last 200 m?"

2. Homework Equations : "(HINT: use V-V0=a*t to find time. Then use X = Vo*t + 0.5*a*t^2 to get t)."

The Attempt at a Solution

:
V-V0=a*t --> 10-6=0.2*t --> 4=0.2*t --> t=4÷0.2=20s.
That self-checks: V-V0=a*t --> V-6=0.2*20 --> V=4+6=10m/s

But the second part is more problematic. X is the distance; I plug in the 20s from the first equation.
X = Vo*t + 0.5*a*t^2[/B]
X = 6*20 +0.5*0.2*(20^2)
X = 120+0.01*400 = 120+40 = 160...instead of 200 as it should be!

It's also possible to put X=200 from the problem, and solve a quadratic for t, but that is a big pain in the rear. Anyway, I am really wondering what silly step I am missing that part 2 gives me 160 instead of 200??!

 

[collinsmark]

Hi Folks: I am puzzled by this problem; there must be something wrong but I can't see what...

1. The problem statement: "A runner enters the last 200 m of a race running 6 m/s. She accelerates at 0.2 m/s2 &crosses the finish line at 10 m/s. What is her time for the last 200 m?"

2. Homework Equations : "(HINT: use V-V0=a*t to find time. Then use X = Vo*t + 0.5*a*t^2 to get t)."

The Attempt at a Solution

:
V-V0=a*t --> 10-6=0.2*t --> 4=0.2*t --> t=4÷0.2=20s.
That self-checks: V-V0=a*t --> V-6=0.2*20 --> V=4+6=10m/s

But the second part is more problematic. X is the distance; I plug in the 20s from the first equation.
X = Vo*t + 0.5*a*t^2
X = 6*20 +0.5*0.2*(20^2)
X = 120+0.01*400 = 120+40 = 160...instead of 200 as it should be!

It's also possible to put X=200 from the problem, and solve a quadratic for t, but that is a big pain in the rear. Anyway, I am really wondering what silly step I am missing that part 2 gives me 160 instead of 200??!
[/B]

The question, as it was written above, is a little vague. If you assume that she starts accelerating at 200 m before the finish line, and continues accelerating until she crosses the finish, the math does not work out and she will cross the finish line at a speed faster than 10 m/s.

One way to interpret the problem such that things work out is that she starts accelerating at some point within the last 200 m of the race, but not necessarily at exactly 200 m before the finish line. If so, the problem statement would be interpreted as, "A runner enters the last 200 m of a race running 6 m/s. At some point within that last 200 m, she accelerates at 0.2 m/s² and continues accelerating until the finish such that she crosses the finish line at 10 m/s.

On the other hand, it's also possible to interpret the problem such that she starts accelerating at the point 200 m before the finish line, and stops accelerating when she reaches 10 m/s, and finishes the remainder of the distance at a constant 10 m/s. That will lead to a different answer than above. And if that's the case, the problem statement would read, "A runner enters the last 200 m of a race running 6 m/s, at which point she accelerates at 0.2 m/s² until she reaches a speed of 10 m/s. She continues at that speed until she crosses the finish line."

I don't know which is the correct way to interpret the problem. Perhaps you can solve it both ways and ask your instructor which is the correct interpretation.

 

[Head_Unit]

Ah, I get it now, thanks. The acceleration was calculated from the velocity change, but the "200 m" stuck on as a fact…but not really calculated. So probably this would be self-consistent if the acceleration was 0.1 m/s^2 or such.