# Question about Estimating with finite Sums

• Sep 14th 2009, 07:24 PM
emonimous
Question about Estimating with finite Sums
Hi guys,
I just started Calculus 2, the topic we're on is estimating with finite sums.
I seem to be having a hard time with this problem in my textbook

Distance from Velocity data

The accompanying table gives data for the velocity of a vintage sports car accelerating from 0 to 142mi/h in 36 seconds(10 thousandths of an hour).

Time..Velocity(mph)
0.0....0
0.001..40
0.002..62
0.003..82
0.004..96
0.005..108
0.006..116
0.007..125
0.008..132
0.009..137
0.01....142

Problem A

Use rectangles to estimate how far the car traveled during the 36 sec it took to reach 142mph

This one I managed to solve pretty easily using using upper and lower sums.

The shortest possible distance :
D= 0.001(0+40+62+82+93+108+116+125+132+137)= .798 mi

The longest possible distance:
D=0.001(40+62+82+96+108+116+125+132+137+142)= 0.94 mi

Problem B

Roughly how many seconds did it take the car to reach the halfway point? About how fast was the car going?

this is the problem I'm having. I'm not sure where to start. I thought I could use halfway point= D/2 then work from there but I'm completely unsure on where to take it from there. A hint in the right direction would be awesome, I'm not looking for someone to solve the problem but feel free anyways.
• Sep 14th 2009, 07:39 PM
emonimous
I might appear crazy for having the one and only reply to my post LOL , but I think I might have it figured out.

What I did is took both upper and lower distance values

\$\displaystyle 0.798/0.01=0.399/t\$
\$\displaystyle 0.94/.01=0.47/t\$

Correct me if I am wrong but I solve for T from there!
• Sep 14th 2009, 07:40 PM
VonNemo19
Quote:

Originally Posted by emonimous
I'm not sure where to start. I thought I could use halfway point= D/2 then work from there but I'm completely unsure on where to take it from there. A hint in the right direction would be awesome, I'm not looking for someone to solve the problem but feel free anyways.

Yeah, use D/2. Then the sum reduces by half of the terms. Insert an unknow for the quantity youre after...

d/2=.001(40+60+82+96+?)

Notice the word..."roughly"...