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 Physics Help!
 JSC_23 From New Zealand, joined Jul 2001, 223 posts, RR: 0Posted Sun Mar 16 2003 10:29:31 UTC (13 years 2 months 1 week 5 days 14 hours ago) and read 1369 times:

 for homework we have this question : "while travelling at 40 ms a driver sees a red light 100m away.Calculate the steady acceleration he must give the car if he is to stop at the light"... i can't work out the formula to use...(or how to rearrange one..)basically the things we are given are velocity initial:40ms velocity final:0 ms distance:100m and i need to find accelleration? any ideas....help would be much appreciated... kindest regards JSC_23
 Be a realist...Remain a dreamer!
 Positive rate From Australia, joined Sep 2001, 2143 posts, RR: 1 Reply 1, posted Sun Mar 16 2003 11:11:18 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1335 times:

 This is very rudimentary in my head calculations here so someone please correct me if i'm wrong. If you're doing 40m/s(144km/h) and you see a red light 100m away, you will need to decelerate at a steady 16m/s in order to stop in time.
 JSC_23 From New Zealand, joined Jul 2001, 223 posts, RR: 0 Reply 2, posted Sun Mar 16 2003 11:13:19 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1334 times:

 Im sorry for this but also I have this question: "how long does it take to hit the ground if you jump off a wall 3m high?" I work out that the acceleration is 10 m/s/s (gravity) so it should work out to about 1/3rd of a second 0.333333333 but it seems to easy? JSC_23
 Be a realist...Remain a dreamer!
 Fritzi From United Arab Emirates, joined Jun 2001, 2763 posts, RR: 2 Reply 3, posted Sun Mar 16 2003 11:15:54 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1331 times:

 Thake the equation: d=((Vi+Vf)/2)*T rewrite it so that you solve for T T=(2/(Vi+Vf))*d T=(2/(40+0))*100 T=(2/(40))*100 T=(0,05)*100 T= 5 m/sec(sq)
 Fritzi From United Arab Emirates, joined Jun 2001, 2763 posts, RR: 2 Reply 4, posted Sun Mar 16 2003 11:19:29 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1324 times:

 T= - 5 m/sec(sq)
 JSC_23 From New Zealand, joined Jul 2001, 223 posts, RR: 0 Reply 5, posted Sun Mar 16 2003 11:24:19 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1319 times:

 Thanks Fritzi -Please excuse me as I'm a beginner but isn't that working out time and not acceleration???? JSC_23
 Be a realist...Remain a dreamer!
 Fritzi From United Arab Emirates, joined Jun 2001, 2763 posts, RR: 2 Reply 6, posted Sun Mar 16 2003 11:28:48 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1316 times:

 q2) d=0,5GT(squared) 3=0,5*-10T(Sq) 3/(5) = T(Sq) (Sq root)0,6= (Sq root)T(Sq) T =0,775 sec I did this one very quickly so there may be some mistakes
 Fritzi From United Arab Emirates, joined Jun 2001, 2763 posts, RR: 2 Reply 7, posted Sun Mar 16 2003 11:30:45 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1311 times:

 The answer for acceleration is positive, the answer for decelleration is negative. Its the same formula.
 Fritzi From United Arab Emirates, joined Jun 2001, 2763 posts, RR: 2 Reply 8, posted Sun Mar 16 2003 11:33:17 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1308 times:

 I may have screwed up but I dont have any more time right now because I have my IB A1 HL Swedish oral in 1.5 hours and I am a bit stressed
 JSC_23 From New Zealand, joined Jul 2001, 223 posts, RR: 0 Reply 9, posted Sun Mar 16 2003 11:34:09 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1306 times:

 im sure your right but i can't understand what the T is for in the first equation....shouldn't it be A=,??? Thanks for you help... JSC_23
 Be a realist...Remain a dreamer!
 Fritzi From United Arab Emirates, joined Jun 2001, 2763 posts, RR: 2 Reply 10, posted Sun Mar 16 2003 11:36:44 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1300 times:

 Vf(Sq)=Vi(Sq)+(2*A*2) may be the equation that you are looking for. Sorry if I mislead you. As I said, I am under a lot of pressure right now.
 SunAir From India, joined Jun 2009, 0 posts, RR: 1 Reply 11, posted Sun Mar 16 2003 11:43:41 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1295 times:

 Uhh, or you just use this equation: v=u + 2as v=0ms u=40ms s=100m 0=40 + 2(a)(100) -40 = 200a a=-5ms^-2 Therefore acceleration = 5ms^-2 deceleration
 JSC_23 From New Zealand, joined Jul 2001, 223 posts, RR: 0 Reply 12, posted Sun Mar 16 2003 11:43:47 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1293 times:

 Please do NOT be sorry........ You are helping me alot....i very much appreciate that!!!!! JSC_23
 Be a realist...Remain a dreamer!
 JSC_23 From New Zealand, joined Jul 2001, 223 posts, RR: 0 Reply 13, posted Sun Mar 16 2003 11:52:46 UTC (13 years 2 months 1 week 5 days 13 hours ago) and read 1283 times:

 BEST OF LUCK for your exam Fritzi...i hope everything goes well!!!!!!!!! JSC_23
 Be a realist...Remain a dreamer!
 Delta-flyer From United States of America, joined Jul 2001, 2682 posts, RR: 6 Reply 14, posted Sun Mar 16 2003 16:03:06 UTC (13 years 2 months 1 week 5 days 8 hours ago) and read 1248 times:

 Sorry, guys, all wrong!!! Let an old fart do it right...... Start with the distance-time-acceleration formula: d = 0.5 * a * t^2 ...EQ(1) d = distance = 100m a = acceleration = ? t = time to decelarate from 40 m/s to 0 = ? ( t^2 = t "squared") Next, the definition of acceleration is rate of change of speed ..... at constant acceleration, it is simply the change in speed divided by the time taken for that change ..... a = (Vf - V0) / t ...EQ(2) Vf = final velovity = 0 m/s V0 = initial velocity = 40 m/s (by the way, velocity is m/s, not ms!) Note that we have velocities and distance, want to solve for acceleration. So let's eliminate time from equation (1) by substituting equation (2), thus .... Rewriting EQ(2): t = (Vf - V0) / a ...EQ(3) Substitute for t in EQ(1): d = 0.5 * a * { (Vf V0) / a }^2 Collect terms, simplify: d = 0.5 * (Vf - V0)^2 / a or, d = (Vf - V0)^2 / (2 * a) Now, rearrange to solve for a: a = (Vf V0)^2 / (2 * d) Finally, plug in the numbers: a = (0 - 40)^2 / (2 * 100) = 1600 / 200 = 8 m/s^2 (metres per second per second) Now, to check: Substitute into equation (3): t = (0 - 40) / 8 = 5 seconds Substitute into equation (1), using t = 5 s: t = 0.5 * 8 * (5)^2 = 0.5 * 8 * 25 = 100 metres It checks out!! ===================== Now, Fritzi was almost there .... he solved for time correctly (5 sec) but inadvertantly called it acceleration ... see above ... he gave T = 5 m/s (sq) should have been 5 sec. ====================== Here's a trick: you can easily determine if an equation is NOT correct if the units don't work out. For example, SunAir gave v=u + 2as (v, u are velocity, a is acceleration, s is distance) - here's how to check it out ..... Substitute units for variables: v = u + 2as m/s = m/s + (2) m/s/s * m {note the constant 2 has no units - just ignore it} Note the last term, when you simplify, becomes m*m /s/s or (m/s)^2 That is velocity squared, and is inconsistent with the other terms which are velocity. This methodology is a part of "dimensional analysis", which is used in some sciences to construct equations where physical models are not well understood. ================= You second problem can also use equation (1), but you have to rewrite it as a function of time: t^2 = d / (0.5 * a) but a = accel of gravity, or g = 10 m/s^2 t^2 = d / (0.5 * g) t = sqrt { d / (0.5 * g) } Plug in numbers ...... t = sqrt { 3 / (5) } = sqrt { 0.6 } = 0.775 seconds Fritzi is correct!!! I hope you can follow this. Please don't just copy it, make sure you understand it. I'll be glad to answer any questions. I'm working on my taxes today, so I'll be on my computer off and on all day. Good luck, Pete
 "In God we trust, everyone else bring data"
 SunAir From India, joined Jun 2009, 0 posts, RR: 1 Reply 15, posted Mon Mar 17 2003 07:55:52 UTC (13 years 2 months 1 week 4 days 16 hours ago) and read 1181 times:

 "This methodology is a part of "dimensional analysis", which is used in some sciences to construct equations where physical models are not well understood." So are these equations (v=u+at, v^2=U^2+2as etc...) incorrect? These equations are part of what I got my final physics school certificate on!
 Delta-flyer From United States of America, joined Jul 2001, 2682 posts, RR: 6 Reply 16, posted Fri Mar 21 2003 04:19:46 UTC (13 years 2 months 1 week 20 hours ago) and read 1131 times:

 So are these equations (v=u+at, v^2=U^2+2as etc...) incorrect? No, the above equations are quite correct. However, you wrote: v=u + 2as which is incorrect, and led you to the wrong answer. My dimensional analysis example also showed that this equation cannot be correct, as the units are inconsistent. Sorry, Pete
 "In God we trust, everyone else bring data"
 Flight152 From United States of America, joined Nov 2000, 3508 posts, RR: 6 Reply 17, posted Fri Mar 21 2003 04:41:40 UTC (13 years 2 months 1 week 20 hours ago) and read 1128 times:

 Where is JetService when you need him?
 Delta-flyer From United States of America, joined Jul 2001, 2682 posts, RR: 6 Reply 18, posted Fri Mar 21 2003 05:10:36 UTC (13 years 2 months 1 week 19 hours ago) and read 1118 times:

 This problem has obviously left him ....... ..... speechless?!!
 "In God we trust, everyone else bring data"
 Redngold From United States of America, joined Mar 2000, 6907 posts, RR: 40 Reply 19, posted Fri Mar 21 2003 05:29:27 UTC (13 years 2 months 1 week 19 hours ago) and read 1116 times:

 NEGATIVE! Remember, if you are decelerating your answer must be a negative number! redngold
 Up, up and away!
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