# Homework Help: Question about Geometry vs. Calculus in Physics Problem

1. Sep 25, 2011

### ocard232

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?

1. The problem statement, all variables and given/known data

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

This graph is given:

2. Relevant equations

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

3. 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.

So, now for my question: Why aren't they the same? Did I make a mistake somewhere, or is my understanding incorrect?

2. Sep 25, 2011

### Staff: Mentor

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.

3. Sep 25, 2011

### ocard232

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