Question about Geometry vs. Calculus in Physics Problem

AI Thread Summary
The discussion revolves around a physics problem involving the calculation of distance traveled by a car using two methods: calculus and geometry. The user initially calculated the distance using definite integrals, arriving at an incorrect total of 45 meters. They then used geometric methods to find the area under the curve, which correctly yielded 55 meters. The confusion stemmed from an incorrect equation for the velocity during the second interval, which was later identified and corrected. Ultimately, the resolution highlighted the importance of ensuring that the equations used match the graphical representation of the problem.
ocard232
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Okay, I'm not asking how to find the solution to this problem. I already found the solution they were looking for. The thing that confuses me is that I got two different solutions using two methods that should have given me the same answer. Could someone show me what I did wrong?

Homework Statement



Find the distance d0,2 traveled by the car between t = 0 s and t = 2 s.

This graph is given:
PhysicsQuestion1.jpg


Homework Equations



Equations:
For t = 0s to t = 1s: V(t) = 30t
For t = 1s to t = 2s: V(t) = 20t

The Attempt at a Solution



The distance is the area under the curve. You should be able to solve this either with calculus or with geometry.

I used calculus first. I took the definite integral of both equations and added them together.

The definite integral from 0 to 1 of 30t is 15. (15(1)^2 - 15(0)^2 = 15)
The definite integral from 1 to 2 of 20t is 30. (10(2)^2 - 10(1)^2 = 30)
Adding them together, the total distance traveled should have been 45 m.

This answer was incorrect. So I tried using geometry.

The first interval was a triangle. The area of a triangle is (1/2)b*h.
For 0 to 1: b = 1, h = 30. (1/2)(1*30) = 15. Same as the definite integral.

The second interval was a triangle on top of a rectangle. The area of a rectangle is b*h.
For 1 to 2: b = 1, hrectangle = 30, htriangle = 20.
(1*30) + (1/2)(1*20) = 40. Not the same as the definite integral.

Adding them together, I got 55, which was the correct answer.

So, now for my question: Why aren't they the same? Did I make a mistake somewhere, or is my understanding incorrect?
 
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ocard232 said:
Equations:
For t = 0s to t = 1s: V(t) = 30t
OK, this one matches the diagram.
For t = 1s to t = 2s: V(t) = 20t
But this one doesn't match up. Check a couple of values, such as t = 1 and t = 2.

Fix that second equation.
 
Doc Al said:
OK, this one matches the diagram.

But this one doesn't match up. Check a couple of values, such as t = 1 and t = 2.

Fix that second equation.

XD Ahhh... so simple. Okay, the equation for the interval is 20t + 10. Thanks.
 
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