# \[Urgent\] Physics Help **URL:** \ **Category:** random\ **Created:** [September 3, 2004, 4:44am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136 "2004-09-03T04:44:17Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![Yeldarb](https://avatars.discourse-cdn.com/v4/letter/y/51bf81/32.png) [@Yeldarb](https://forum.kirupa.com/u/Yeldarb)\ **Post date:** [September 3, 2004, 4:44am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/1 "2004-09-03T04:44:17Z")
I know there are some pretty smart people browsing these forums, and I need a bit of quick help with physics. I’m working on an assignment for my AP Physics class through webassign, which is basically a way that teachers can put assignments online for students and they can fill out the homework on the internet and submit it for instant grading. Unfortunately they can also set the time it is due, and mine happens to be due in about 15 minutes. I have been working on the last problem for about an hour and a half, and just can’t figure it out. If you would be so kind to explain it to me, I would be very grateful. Question: > A bicyclist is finishing his repair of a flat tire when a friend rides by at [color=#dd0000]3.8[/color] m/s. Two seconds later, the bicyclist hops on his bike and accelerates at [color=#dd0000]2.0[/color] m/s2 until he catches his friend. Here’s the equation I have so far: > [font=Arial, Helvetica, sans-serif]v01t+.5a1t2=v02t+.5a2t2[/font] But that doesn’t facter in the 2 second head start. Thanks. ---
**Author:** ![system](https://canada1.discourse-cdn.com/flex011/uploads/kirupa/original/3X/6/2/621e5c11736f46532526e61be85940af4230f3e5.png) [@system](https://forum.kirupa.com/u/system)\ **Post date:** [September 3, 2004, 5:01am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/2 "2004-09-03T05:01:12Z")
Nevermind about the urgent part, the deadline has passed now, but I would still appreciate if someone could explain this to me so I will know how to do problems like this in the future. ---
**Author:** ![system](https://canada1.discourse-cdn.com/flex011/uploads/kirupa/original/3X/6/2/621e5c11736f46532526e61be85940af4230f3e5.png) [@system](https://forum.kirupa.com/u/system)\ **Post date:** [September 3, 2004, 6:08am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/3 "2004-09-03T06:08:42Z")
umm…carry the 2 and add 6…just kiddin, I am taking physics next year, and after reading that…I’m not to sure if I will now… ---
**Author:** ![system](https://canada1.discourse-cdn.com/flex011/uploads/kirupa/original/3X/6/2/621e5c11736f46532526e61be85940af4230f3e5.png) [@system](https://forum.kirupa.com/u/system)\ **Post date:** [September 3, 2004, 6:19am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/4 "2004-09-03T06:19:25Z")
> A bicyclist is finishing his repair of a flat tire when a friend rides by at 3.8 m/s. Two seconds later, the bicyclist hops on his bike and accelerates at 2.0 m/s2 until he catches his friend. What is the question??? The “question” you have posted is a statement Cheers Brownie ---
**Author:** ![system](https://canada1.discourse-cdn.com/flex011/uploads/kirupa/original/3X/6/2/621e5c11736f46532526e61be85940af4230f3e5.png) [@system](https://forum.kirupa.com/u/system)\ **Post date:** [September 3, 2004, 8:15am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/5 "2004-09-03T08:15:40Z")
(2sec \* 3.8m/s) + (3.8m/s \* t) = 1/2 \* 2m/s/s \* t^2; solves for the time taken; and solve the first half for the distance required. ---
**Author:** ![system](https://canada1.discourse-cdn.com/flex011/uploads/kirupa/original/3X/6/2/621e5c11736f46532526e61be85940af4230f3e5.png) [@system](https://forum.kirupa.com/u/system)\ **Post date:** [September 3, 2004, 8:48am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/6 "2004-09-03T08:48:56Z")
> [@Yeldarb](#): > > Here’s the equation I have so far: > v01t+.5a1t²=v02t+.5a2t² > But that doesn’t facter in the 2 second head start. Well, after time t, both bikes have travelled the same distance D. After time T, bike 1 has travelled D = v01t+.5a1t² After time T, bike 2 has travelled D = v02t+.5a2t² + 2v02 (he’s travelled 2 secondes at the speed of v02 hence the 2v02). BUT when bike 1 starts, it’s speed it null, so v01 = 0, and since there’s no mention of bike 2 excelerating, we can assume a02 = 0 SO: v01t+.5a1t² = v02t+.5a2t² + 2v02 .5a1t² = v02t+ 2v02 a1t² - 2v02t + 4v02 = 0 (what ciab said) equation of the second degree =\> piece of cake! ---
**Author:** ![system](https://canada1.discourse-cdn.com/flex011/uploads/kirupa/original/3X/6/2/621e5c11736f46532526e61be85940af4230f3e5.png) [@system](https://forum.kirupa.com/u/system)\ **Post date:** [September 3, 2004, 11:53am UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/7 "2004-09-03T11:53:01Z")
alright, thanks, I get it now 😃 also, sorry about forgetting the question, you guys assumed right, they wanted to know how long it would take for the two guys to be at the same distance away. ---
**Author:** ![system](https://canada1.discourse-cdn.com/flex011/uploads/kirupa/original/3X/6/2/621e5c11736f46532526e61be85940af4230f3e5.png) [@system](https://forum.kirupa.com/u/system)\ **Post date:** [September 3, 2004, 5:49pm UTC](https://forum.kirupa.com/t/urgent-physics-help/63136/8 "2004-09-03T17:49:57Z")
> [@Ilyas Premier](#): > > Well, after time t, both bikes have travelled the same distance D. > > After time T, bike 1 has travelled D = v01t+.5a1t² > After time T, bike 2 has travelled D = v02t+.5a2t² + 2v02 (he’s travelled 2 secondes at the speed of v02 hence the 2v02). > > BUT when bike 1 starts, it’s speed it null, so v01 = 0, and since there’s no mention of bike 2 excelerating, we can assume a02 = 0 SO: > > v01t+.5a1t² = v02t+.5a2t² + 2v02 > .5a1t² = v02t+ 2v02 > a1t² - 2v02t + 4v02 = 0 (what ciab said) > > equation of the second degree =\> piece of cake! _Twitch_ Thank god I got done with the class.