Solving Algebra Roadblock: 100m Dash Force & Speed

[jared69sib]

Algebra Road Block...

Homework Statement

The 100m dash can be run by the best sprinters in 10 seconds. A 66-kg sprinter accelerates uniformly for the first 45m to reach top speed, which he maintains for the remaining 55 meters. a) What is the avg horizontal component of force exerted on his feet by the ground during acceleration? b) What is the speed of the sprinter over the last 55m of the race (ie., his top speed)?

Homework Equations

[URL]http://24.45.232.22:8181/all/TB_SCHOOL/PHY130/ch4_25.JPG[/URL]

The Attempt at a Solution

How did they get t1 on the 3rd line?

t1 = 20d1 / (d2 + 2d1)

I am just NOT seeing it...

Thanks in advance! :)

 

[supratim1]

let acceleration be a, final constant velocity be v, and time taken to reach that velocity be t.

therefore, 45 = 1/2 a.t^2

and 55 = v(10-t)

also, v = at.

solve and get a and v.

F = ma.

 

[jared69sib]

let acceleration be a, final constant velocity be v, and time taken to reach that velocity be t.

therefore, 45 = 1/2 a.t^2

and 55 = v(10-t)

also, v = at.

solve and get a and v.

F = ma.

Sorry. This is where I am stuck... I don't know how to solve for 2 variables at once. I know I should have just as many equations as variables, but my algebra skills are horrible...

 

[supratim1]

then i advise you personally learn it from a teacher as soon as possible. it would be a thousand times better learning experience than if i explain you here.

 

[jared69sib]

then i advise you personally learn it from a teacher as soon as possible. it would be a thousand times better learning experience than if i explain you here.

I was specific with my question. Don't reply unless you can answer my question. FYI, if I could get a teachers help don't you think I would have?

Can anyone actually help me?

 

[eumyang]

I think that was supratim1's point. If your algebra skills are horrible, you really need to review them ASAP. What you're asking is about literal equations, an Algebra 1 topic (if you're in the US).

I'll show the first step and tell you the rest, but try to work it out yourself. From this:
d_2 t_1 = 2d_1 (10.0 - t_1)

... distribute the RHS (right-hand side).
d_2 t_1 = 20.0d_1 - 2d_1 t_1

Add the 2d₁t₁ term to both sides. Factor out the t₁ on the LHS. Divide both sides by the expression in parentheses on the LHS. You will get:
t_1 = \frac{20.0d_1}{d_2 + 2d_1}

 

[jared69sib]

I think that was supratim1's point. If your algebra skills are horrible, you really need to review them ASAP. What you're asking is about literal equations, an Algebra 1 topic (if you're in the US).

I'll show the first step and tell you the rest, but try to work it out yourself. From this:
d_2 t_1 = 2d_1 (10.0 - t_1)

... distribute the RHS (right-hand side).
d_2 t_1 = 20.0d_1 - 2d_1 t_1

Add the 2d₁t₁ term to both sides. Factor out the t₁ on the LHS. Divide both sides by the expression in parentheses on the LHS. You will get:
t_1 = \frac{20.0d_1}{d_2 + 2d_1}

Thank you very much. That's all I wanted to know. Occasionally, I become blinded by my thoughts, and a simple thing is just overseen. I'm taking differential equations and physics for scientists and engineers right now, so I don't have a teacher willing to teach me "basics". Seeing an intermediate step makes all the difference.

Thank you! :)

 

[supratim1]

'let acceleration be a, final constant velocity be v, and time taken to reach that velocity be t.

therefore, 45 = 1/2 a.t^2

and 55 = v(10-t)

also, v = at.

solve and get a and v.

F = ma.'
that was enough of help and answer. rest is your job! :-)

 

[jared69sib]

'let acceleration be a, final constant velocity be v, and time taken to reach that velocity be t.

therefore, 45 = 1/2 a.t^2

and 55 = v(10-t)

also, v = at.

solve and get a and v.

F = ma.'



that was enough of help and answer. rest is your job! :-)

I don't wish to argue with you. What does the title say? "Algebra road block". The physics and calculus part is easy. I know what to do. Thanks for the effort, but the standardized question format of this website should have prevented this problem. Next time stop thinking of how "stupid" I might be and finish reading the post.