How Far Did the Drag Racer Travel?

[dvsj67]

Homework Statement

A drag racer’s acceleration is 26.6 meters/second2, and at the end of the race, its final speed is 146.3 meters per second. What is the total distance the drag racer traveled?

Homework Equations

Vf^2=1/2(26.6)(146.3)^2= 409 meters

The Attempt at a Solution

Ok..
The attempt at the solution is my problem.. if any of you are familiar with physics for dummies workbook and read this question on p.34 it says that the equation above = 409 meters.. My questions is how?!

When i enter 1/2(26.6)(146.3)^2 into my calculator i get 284669.077.. Which is a long way off from the answer the book gives.

I am not currently a student in a math program I just find math fascinating but this has been bugging me for weeks. I was hoping someone could show me what i am missing

 

[chroot]

dvsj67,

Welcome to physicsforums!

Could you write down an expression for the car's position, s, as a function of time, t? That would be the best place to begin this problem.

- Warren

 

[dvsj67]

Thank You, Warren

Actually, s is what I am trying to solve i guess it would be. the formula according to the book s=1/2a Vf^2 = Vf^2 = 1/2(26.6)(146.3)^2

So,

s= ?
a= 26.6
Vf^2 = 146.3

 

[gneill]

Thank You, Warren

Actually, s is what I am trying to solve i guess it would be. the formula according to the book s=1/2a Vf^2 = Vf^2 = 1/2(26.6)(146.3)^2

So,

s= ?
a= 26.6
Vf^2 = 146.3

I think there's a parsing issue with the equation. It should read:
$$s = \frac{1}{2 a} v_f^2$$

It derives from the "standard" kinematic equation,
$$v_f^2 - v_i^2 = 2 a s$$
when v_i = 0.

The numerics are dodgy, too. The result should be closer to 402m, not 409m.

 

[dvsj67]

Yes thank you.. that's how it looks.. but between 402m and 409m i can't get anywhere near those numbers

I am just trying to understand how it equals that

 

[gneill]

Yes thank you.. that's how it looks.. but between 402m and 409m i can't get anywhere near those numbers

I am just trying to understand how it equals that

You must be suffering calculation order-of-operation issues. What do get for v²? (show us). Next divide that by 2. What have you got so far?. Next divide that by a. What's the result?

 

[chroot]

Yes, I think there's an error in the book, or you've copied the equation down incorrectly.

Take a close look at the equation you posted, $s = \frac{1}{2} a v_f^2$. The units don't work out -- meters/second^2 times meters^2/seconds^2 certainly does not come out to be meters.

I always start these kinds of problems in one of two places. If I'm given a value for time, then I use this equation, which is straightforward:

$$s(t) = s_0 + v_0 t + \frac{1}{2} a t^2$$

If I'm not given a value for time, then I start with Torricelli's equation, which can be derived from the above. It's independent of time, so makes things a little easier:

$$v_f^2 = v_i^2 + 2 a s$$

In this case, I would start with Torricelli's.

- Warren

 

[dvsj67]

I believe v^2 is already accounted for as 146.3

so if i went 146.3/ 26.6 i get 5.5

The whole question and answer is already written out in the book.
What i am not understanding is how the book got the answer it did..

If i enter it in a calculator i enter 1 Divided by 2 multiplied by 26.6 multiplied by 146.3^2 and i end up with a huge number..

I am debating making a video of what i am seeing and doing since i am not familiar with how to make the formulas on here like gneill

 

[chroot]

dvsj67, if I rearrange Torricelli's equation to solve for position, s, I can get this expression:

$$s = \frac{v_f^2}{2 a}$$

If I plug in 146.3 m/s for $v_f$, and 26.6 m/s^2 for a, I get s = 402.3 meters.

As gneill suggested, to enter this in a calculator, you must be careful with the order of operations. You could do the following:

146.3 * 146.3 = 21403.69
answer / 2 = 10701.845
answer / 26.6 = 402.325

You may wish to use a more sophisticated calculator that supports parentheses, or even https://www.google.com/search?gcx=c...,cf.osb&fp=7ccfaba8b72cb7e1&biw=1358&bih=739"!

- Warren

 

[dvsj67]

Thank You, Warren

That actually clarified this for me, enough that after a week of writing and rewriting this out I am confident to try a similar question