How Do You Calculate Acceleration and Time for a Sprinter Reaching Top Speed?

[cjm1994]

Homework Statement

A sprinter reaches his top speed of 11m/s in t seconds from rest with essentially constant acceleration. If the maintains this speed once reached and covers the 100m distance in 11.5s, Find:

a) the time it took to accelerate.

b) his acceleration until he reaches constant speed.

Homework Equations

Vf^2 = Vi^2 + 2ad
d = Vi(t) + 1/2at^2
Vf = Vi +at

The Attempt at a Solution

I can't figure out how to do this without a least one more variable given.

If I try to use Vf^2 = Vi^2 + 2ad I am missing the acceleration and the distance it accelerates for.
If I try to use d = Vi(t) + 1/2at^2 I am missing the distance, acceleration, and the time.
If I try to use Vf = Vi +at I am missing the acceleration and the time.

I don't really know where to even start, I can't find any questions online that are the same as this one. They usually give you the distance traveled before acceleration stops or how long it took. This is very confusing to me.

 

[Chestermiller]

Hi cgm1994. Welcome to Physics Forums!
If he reaches a speed of 11m/s in t seconds from rest with essentially constant acceleration, what is his acceleration a in terms of t? If you substitute this into your second Relevant Equation, what do you get for his distance covered during the time interval from t = 0 to t=t? How much time is left to run at constant 11 m/s for the rest of the 11.5 sec (in terms of t)? How much distance does he cover during this time interval?

Chet

 

[cjm1994]

Hi cgm1994. Welcome to Physics Forums!
If he reaches a speed of 11m/s in t seconds from rest with essentially constant acceleration, what is his acceleration a in terms of t? If you substitute this into your second Relevant Equation, what do you get for his distance covered during the time interval from t = 0 to t=t? How much time is left to run at constant 11 m/s for the rest of the 11.5 sec (in terms of t)? How much distance does he cover during this time interval?

Chet

Thanks for the reply. I'm not entirely sure what "a in terms of t" means. I understand the rest of your post.

 

[Chestermiller]

Thanks for the reply. I'm not entirely sure what "a in terms of t" means. I understand the rest of your post.

a = 11/t

Chet

 

[cjm1994]

Thats what I though you meant, but that just makes me get d = 5.5t

I'm sorry if I'm just missing something here my physics is really rusty

 

[Chestermiller]

Thats what I though you meant, but that just makes me get d = 5.5t

I'm sorry if I'm just missing something here my physics is really rusty

Yes. 5.5t is correct for the distance traveled by the sprinter while he is accelerating. Now, in terms of t, how far does he travel during the 11.5-t seconds that he is not accelerating?

Chet

 

[cjm1994]

Ok assuming I did this part right I think I got the rest of the question down.

11m/s * (11.5s - t) = 100 - (5.5t)
so
t = 4.81818181

Is that right? Thanks again for the help.

 

[Chestermiller]

Ok assuming I did this part right I think I got the rest of the question down.

11m/s * (11.5s - t) = 100 - (5.5t)
so
t = 4.81818181

Is that right? Thanks again for the help.

Looks right.

Chet