my friend has a physics problem.. - GoFuckYourself.com

robfantasy · 05-11-2004, 03:34 PM

driving along a freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 8 mi/h , the time to go one mile decreases by 10 s . What was your original speed?

how do u get this answer..

SicChild · 05-11-2004, 03:42 PM

I think it's a system problem.

v = d / t

so,

t = 1 (1 mile) / v
and
t - 10 = 1 / (v + 8)

isolate a variable and solve for it...

ADL Colin · 05-11-2004, 04:36 PM

v1 = first velocity
v2= second velocity
d1- first distance
d2 = second distance

d1 = d2 = 1
Since d = v/t

v1*t = v2*t2 = 1

Using:
v2 = v1 +8
and t2 = t = 10

(v1 + 8)(t-10) = 1
10v1 ? 8t + 80 = 0
10v1^2 + 80v1 ? 8 = 0

Using quadratic formula with a=10, b=80, c = -8

You get v1 = ~ .09878 mph which you can verify is correct by :::

In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds

peteinoz · 05-11-2004, 04:40 PM

Quote:

Originally posted by Colin
v1 = first velocity
v2= second velocity
d1- first distance
d2 = second distance

d1 = d2 = 1
Since d = v/t

v1*t = v2*t2 = 1

Using:
v2 = v1 +8
and t2 = t = 10

(v1 + 8)(t-10) = 1
10v1 ? 8t + 80 = 0
10v1^2 + 80v1 ? 8 = 0

Using quadratic formula with a=10, b=80, c = -8

You get v1 = ~ .09878 mph which you can verify is correct by :::

In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds

Erhm YEah.. what "HE" said

Holly · 05-11-2004, 06:03 PM

Quote:

Originally posted by Colin
v1 = first velocity
v2= second velocity
d1- first distance
d2 = second distance

d1 = d2 = 1
Since d = v/t

v1*t = v2*t2 = 1

Using:
v2 = v1 +8
and t2 = t = 10

(v1 + 8)(t-10) = 1
10v1 ? 8t + 80 = 0
10v1^2 + 80v1 ? 8 = 0

Using quadratic formula with a=10, b=80, c = -8

You get v1 = ~ .09878 mph which you can verify is correct by :::

In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds

freeadultcontent · 05-11-2004, 06:06 PM

Thank you for reminding me I suck at all math asside from accounting.

KDizzla · 05-11-2004, 07:49 PM

Quote:

Originally posted by freeadultcontent
Thank you for reminding me I suck at all math asside from accounting.

Math has always been a bad to me.

Holly · 05-11-2004, 07:56 PM

Quote:

Originally posted by KDizzla
Math has always been a bad to me.

I just noticed that in the Jessica Lynch thread.

ADL_Christy · 05-11-2004, 08:01 PM

I knew Colin would have the answer to this!!!!

ADL Colin · 05-12-2004, 04:49 AM

Quote:

Originally posted by ADL_Christy
I knew Colin would have the answer to this!!!!

Hey Christy! :-)

hjnet · 05-12-2004, 05:35 AM

Quote:

Originally posted by Colin
v1 = first velocity
v2= second velocity
d1- first distance
d2 = second distance

d1 = d2 = 1
Since d = v/t

v1*t = v2*t2 = 1

Using:
v2 = v1 +8
and t2 = t = 10

(v1 + 8)(t-10) = 1
10v1 ? 8t + 80 = 0
10v1^2 + 80v1 ? 8 = 0

Using quadratic formula with a=10, b=80, c = -8

You get v1 = ~ .09878 mph which you can verify is correct by :::

In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds

I don't wanna be the ass here, but I think you mixed up seconds and hours somewhere.

At the beginning he drove with 50mph (and needed 72 sec for 1mile), then he increased his speed to 58mph and therefore only needed 62 sec for a mile

ADL Colin · 05-12-2004, 05:51 AM

Quote:

Originally posted by hjnet
I don't wanna be the ass here, but I think you mixed up seconds and hours somewhere.

At the beginning he drove with 50mph (and needed 72 sec for 1mile), then he increased his speed to 58mph and therefore only needed 62 sec for a mile

Haha. Oh, shit. You're right. Forgot to convert the units.

05-11-2004, 03:34 PM	#1
robfantasy Confirmed User Industry Role: Join Date: Jun 2002 Location: Medellin, Colombia Posts: 6,445	my friend has a physics problem.. driving along a freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 8 mi/h , the time to go one mile decreases by 10 s . What was your original speed? how do u get this answer.. __________________ Looking to speak w/ high volume nutra CPA affiliates or networks... msg me

05-11-2004, 04:36 PM	#3
ADL Colin Too lazy to set a custom title Join Date: Feb 2001 Location: Tube Titans, USA Posts: 11,929	v1 = first velocity v2= second velocity d1- first distance d2 = second distance d1 = d2 = 1 Since d = v/t v1t = v2t2 = 1 Using: v2 = v1 +8 and t2 = t = 10 (v1 + 8)(t-10) = 1 10v1 ? 8t + 80 = 0 10v1^2 + 80v1 ? 8 = 0 Using quadratic formula with a=10, b=80, c = -8 You get v1 = ~ .09878 mph which you can verify is correct by ::: In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds __________________ Adult Date Link - $50 PPS starting NOW! -- good and JUICY! skype = "adultdatelink"

05-11-2004, 06:06 PM	#6
freeadultcontent Confirmed User Join Date: Oct 2002 Location: "evitcepsrep ruoy egnahc" Posts: 9,976	Thank you for reminding me I suck at all math asside from accounting. __________________ Exclusive Ethnic & Microniche Sites

05-11-2004, 08:01 PM	#9
ADL_Christy Confirmed User Join Date: Mar 2004 Posts: 304	I knew Colin would have the answer to this!!!! __________________ Adult Video Dating Affiliate Program Hosted Galleries - 50/50 Partner Program 3-Tier Webmaster Referral Program 10%/4%

05-11-2004, 03:42 PM	#2
SicChild Confirmed User Join Date: Mar 2003 Posts: 365	I think it's a system problem. v = d / t so, t = 1 (1 mile) / v and t - 10 = 1 / (v + 8) isolate a variable and solve for it...