Infinity Question (some physics background)

[PatrickDinehart]

I have a noobish question but I haven't been able to figure it out. In physics, we did a design lab (make a lab up on the spot and do it) and I was trying to calculate the correlation between the angle of a ramp to the acceleration of a car that went down the ramp. Of course, we are at the beginning of the year so we're limited to simple kinematics and derivatives/integrals. At a ramp of 0 degrees, the time for the car to go down it is infinite and thus, the acceleration would be calculated under the following premises:

X=x0 + vot + (1/2)a(t^2)

To make this plug able into Excel, I moved a (acceleration) to one side of the equation and plugged in x, x0, v0, and t. The problem here is t is infinite. For reference, the equation ended up being: =0.5*((0+0*(C8)+(C8^2))/1.0385) where C8 is the AVG Time (5 trials 5 times ect).

Is it possible I have used the wrong infinity symbol or is this simply incalculable?

Thanks in advance,
Patrick

 

[ranman256]

I dont think the infinity symbol has a value in Excel.

 

[shg]

The car doesn't take an infinite amount of time to go down the ramp when the slope is zero; the time increases without limit ('approaches infinity') as the slope of the ramp decreases toward zero:

Programming languages don't know about infinity.

 

[shg]

The acceleration of the car is the acceleration of gravity times the sine of the slope:

[Table="width:, class:grid"][tr][td] [/td][td]

A​

[/td][td]

B​

[/td][td]

C​

[/td][td]

D​

[/td][/tr]
[tr][td]

1​

[/td][td]

Slope [deg]​

[/td][td]

-5​

[/td][td][/td][td]B1: Input[/td][/tr]

[tr][td]

2​

[/td][td]

accel [m/s/s]​

[/td][td]

-0.854​

[/td][td][/td][td]B2: =9.8*SIN(RADIANS(B1))[/td][/tr]

[tr][td]

3​

[/td][td][/td][td][/td][td][/td][td][/td][/tr]

[tr][td]

4​

[/td][td][/td][td]

v0 [m/s]​

[/td][td]

x0 [m/s]​

[/td][td][/td][/tr]

[tr][td]

5​

[/td][td][/td][td]

0​

[/td][td]

0​

[/td][td][/td][/tr]

[tr][td]

6​

[/td][td]

t​

[/td][td]

v [m/s]​

[/td][td]

x [m]​

[/td][td][/td][/tr]

[tr][td]

7​

[/td][td]

0​

[/td][td]

0.00​

[/td][td]

0.00​

[/td][td]B7: =$B$2*A7[/td][/tr]

[tr][td]

8​

[/td][td]

1​

[/td][td]

-0.85​

[/td][td]

-0.43​

[/td][td]C7: =$C$5 + $B$5*A7 + $B$2*A7^2/2[/td][/tr]

[tr][td]

9​

[/td][td]

2​

[/td][td]

-1.71​

[/td][td]

-1.71​

[/td][td][/td][/tr]

[tr][td]

10​

[/td][td]

3​

[/td][td]

-2.56​

[/td][td]

-3.84​

[/td][td][/td][/tr]

[tr][td]

11​

[/td][td]

4​

[/td][td]

-3.42​

[/td][td]

-6.83​

[/td][td][/td][/tr]

[tr][td]

12​

[/td][td]

5​

[/td][td]

-4.27​

[/td][td]

-10.68​

[/td][td][/td][/tr]

[tr][td]

13​

[/td][td]

6​

[/td][td]

-5.12​

[/td][td]

-15.37​

[/td][td][/td][/tr]

[tr][td]

14​

[/td][td]

7​

[/td][td]

-5.98​

[/td][td]

-20.93​

[/td][td][/td][/tr]

[tr][td]

15​

[/td][td]

8​

[/td][td]

-6.83​

[/td][td]

-27.33​

[/td][td][/td][/tr]
[/table]

 

[PatrickDinehart]

The acceleration of the car is the acceleration of gravity times the sine of the slope:

[TABLE="class: grid"]
<tbody>[TR]
[TD][/TD]
[TD]

A​

[/TD]
[TD]

B​

[/TD]
[TD]

C​

[/TD]
[TD]

D​

[/TD]
[/TR]
[TR]
[TD]

1​

[/TD]
[TD]

Slope [deg]​

[/TD]
[TD]

-5​

[/TD]
[TD][/TD]
[TD]B1: Input[/TD]
[/TR]
[TR]
[TD]

2​

[/TD]
[TD]

accel [m/s/s]​

[/TD]
[TD]

-0.854​

[/TD]
[TD][/TD]
[TD]B2: =9.8*SIN(RADIANS(B1))[/TD]
[/TR]
[TR]
[TD]

3​

[/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

4​

[/TD]
[TD][/TD]
[TD]

v0 [m/s]​

[/TD]
[TD]

x0 [m/s]​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

5​

[/TD]
[TD][/TD]
[TD]

0​

[/TD]
[TD]

0​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

6​

[/TD]
[TD]

t​

[/TD]
[TD]

v [m/s]​

[/TD]
[TD]

x [m]​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

7​

[/TD]
[TD]

0​

[/TD]
[TD]

0.00​

[/TD]
[TD]

0.00​

[/TD]
[TD]B7: =$B$2*A7[/TD]
[/TR]
[TR]
[TD]

8​

[/TD]
[TD]

1​

[/TD]
[TD]

-0.85​

[/TD]
[TD]

-0.43​

[/TD]
[TD]C7: =$C$5 + $B$5*A7 + $B$2*A7^2/2[/TD]
[/TR]
[TR]
[TD]

9​

[/TD]
[TD]

2​

[/TD]
[TD]

-1.71​

[/TD]
[TD]

-1.71​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

10​

[/TD]
[TD]

3​

[/TD]
[TD]

-2.56​

[/TD]
[TD]

-3.84​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

11​

[/TD]
[TD]

4​

[/TD]
[TD]

-3.42​

[/TD]
[TD]

-6.83​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

12​

[/TD]
[TD]

5​

[/TD]
[TD]

-4.27​

[/TD]
[TD]

-10.68​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

13​

[/TD]
[TD]

6​

[/TD]
[TD]

-5.12​

[/TD]
[TD]

-15.37​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

14​

[/TD]
[TD]

7​

[/TD]
[TD]

-5.98​

[/TD]
[TD]

-20.93​

[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]

15​

[/TD]
[TD]

8​

[/TD]
[TD]

-6.83​

[/TD]
[TD]

-27.33​

[/TD]
[TD][/TD]
[/TR]
</tbody>[/TABLE]

Thanks for the replies so far! On a second note, can you explain the table? We don't necessarily use slopes on the understanding that delta(velocity, position, or acceleration) will be the slope in question.

 

[shg]

The slope of the ramp determines the accelerating force on the car, no?

EDIT: Describe your experiment; what you do, what you record, and what result you're trying to calculate.

 

[PatrickDinehart]

The slope of the ramp determines the accelerating force on the car, no?

EDIT: Describe your experiment; what you do, what you record, and what result you're trying to calculate.

To your first question: Yes, although we can't just say "slope is 3 therefore A must be 3". We have to use kinematics to prove it. Why, I don't know but thus is life. After looking at the chart, I understand it. It took me a minute because of curriculum changes over the last year :D Thanks again!

 

[shg]

So -- sorted?