# Physics Kinematic Question

1. Mar 4, 2009

### DallasStar

1. The problem statement, all variables and given/known data

A runner wants to run a 1500m race with an average velocity of 6.00m/s. After running the first 500 m, he notices that the elapsed time is 1 min 40 seconds. Panicking, he accelerates for "t" seconds at 0.20 m/s² and then finishes the race at his new faster speed. He makes it just in time. How long did he accelerate for? (Use 5 sig. figs. in your answer)

Hint: The answer is between 8.333333333 and 10; whereby 8.3333333 is to short and 10 to long

3. The attempt at a solution

It is easy to figure out that the runner needs 150 sec for the rest of the race (1000m). I figured out that his average velocity for the first part was 5 m/s. However, where I'm stuck is, is when he starts to accelerate what is his initial velocity and his final velocity? I got 5.3333333333 m/s as an initial velocity as well as 5 m/s and as a final velocity i got 6 m/s and 6.6666666666 m/s. (6=(vi+vf)/2)
I am pretty sure that I am "overthinking" this question but I am completely stuck.

Can somebody, please, help me !!!!!!!!

Thank you,

Christian

2. Mar 4, 2009

### S_Happens

You are overthinking it. You have already calculated the "initial velocity" you are looking for. Your major concern is going to be when he is in the 500m to 1500m distance, so the initial velocity is velocity at 500m. The intent of this problem is to have average velocity from 0-500m be your instantaneous at 500m regardless of how that had to happen (no acceleration is mentioned so all you can do is use average).

If his average velocity for the first third of the race is 5m/s, then he must average 6.5m/s over the last 2/3 of the race to have an overall 6m/s average for the whole race. See if you can use this information and the acceleration you are given to figure out the final answer.

Hint: There are two components to the final part of the race.

3. Mar 5, 2009

### DallasStar

Thank you, I understand now that his average velocity for the second part is 6.5 m/s, however when I am trying to calculate the distance he needs to accelerate as well as the time I get 7.5 s as an answer??? What am I doing wrong?

4. Mar 5, 2009

### DallasStar

After several tries i finally figured out that 6.5 m/s is not the final velocity. The final velocity would be more than 6.5 since the runner needs time to accelerate from vi= 5 m/s; however how can I figure out the final velocity? which formula can I use?

Thank you,
Christian

5. Mar 5, 2009

### LowlyPion

You have 3 unknowns effectively.

The distance he accelerates for. And the time he accelerates. (Though there is a simple relationship between these two since you are given the acceleration.)

And the final velocity for the remainder of the run.

You can write one equation for these two readily enough, from the velocity, acceleration and distance equation.

The other may be a little less obvious.

You have already figured that there is 1000 m to cover in the final 150 seconds.

Think about what comprises that distance? There is the acceleration distance and the remaining constant velocity distance and they both need to add to 1000. Happily time is constrained between these two distances because they must add to 150.

Try constructing the distance equation that adds to 1000m. Then you can solve by starting to eliminate unknowns.

6. Mar 5, 2009

### DallasStar

Thank you for your help. I got three equations now for the final velocity --> vf = 5+0.2t; for the displacement he accelerates for --> d = 5t+0.1t²; and for the displacement after he accelerated --> d = 5t + 0.2t². The next step i did was adding the two displacement equations because they should equal 1000; however, when you put that into your quadratic formula you solve for t? and thats where i got lost :(

If i did something wrong please correct me

thank you,

christian

7. Mar 6, 2009

### DallasStar

I am so confused

8. Mar 6, 2009

### alphysicist

Hi DallasStar,

I don't think this last equation is correct. You seem to be multiplying the times together to get the t2; however they are two different times. You have a time t1 for the accelerating part of the motion, and a time t2 for the last part of the race; this last equation will have both times in it. Then you can eliminate t2 because you know what the sum of these times must be, and solve for t1.

9. Mar 6, 2009

### LowlyPion

It is a more complex calculation, so you need to take care constructing your equation for distances.

Maybe examine that in 2 parts to keep it simpler?

Your first distance equation will look like

d = Vo*t + 1/2*a*t2 = 5*t + .1*t2 which you apparently got.

The second part of the distance, the constant velocity phase, is Vf*(150 - t), where t is the time for accelerating, because after all your total time budget is 150 sec.

Now you know both of those distances together = 1000m. Since you can eliminate t fairly easily by noting that Vf = Vo + a*t and putting t in terms of V as t = (Vf - 5)/a , then you will end in a quadratic of Vf.

10. Mar 6, 2009

### DallasStar

I understand the first part, but there you lost me :(. can somebody explain it in simpler words please? i feel like im menatlly challenged right now :(

11. Mar 6, 2009

### LowlyPion

You should have an equation that equals 1000 m expressed in v and t as your only variables.

Why don't you post that?

12. Mar 6, 2009

### LowlyPion

The object then once you have your equation is to solve for your unknowns. So use substitution. You can do that directly by observing that the increased velocity must be given by:

Vf = Vi + a*t

This means that you can relate Vf directly to t.

So in your distance equation substitute everywhere for t

t = (Vf - 5)/.2

Or alternatively substitute everywhere for V the expression

V = 5 + .2*t
(Substituting with this might even be easier.)

Either way you will end with a quadratic in V or t that you need to resolve. So resolve it.