Help With This Physics - Education - Nairaland
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| Help With This Physics by seedord247(op): 4:38am On Oct 14, 2012 |
A train is traveling down a straight track at 29 m/s when the engineer applies the brakes, resulting in an acceleration of −1.0 m/s2 as long as the train is in motion. How far does the train move during a 58-s time interval starting at the instant the brakes are applied? urgent answer needed... |
| Re: Help With This Physics by ATMC(f): 5:01am On Oct 14, 2012 |
5.8m/s |
| Re: Help With This Physics by seedord247(op): 5:08am On Oct 14, 2012 |
ATMC: 5.8m/show sure are you? |
| Re: Help With This Physics by ojotu4real(m): 7:34am On Oct 14, 2012 |
s= ut + 1/2 at*sq. Ur ans wil be zero. Meaning d train would av stopped b4 dat time |
| Re: Help With This Physics by Emmy3(m): 11:46am On Oct 15, 2012 |
S=ut + 1/2at^2 is the total distance covered by the train. . . . S=ut - 1/2at^2 is the distance coverd after the brake was applied. . . But if you are give a -ve acceleration i.e decceleration use the first one cuz it wil automatically change back the sign or use the 2nd one without observing the -ve sign in the retardation. . . |
| Re: Help With This Physics by Emmy3(m): 11:56am On Oct 15, 2012 |
ATMC: 5.8m/snot velocity bt distance is required |
| Re: Help With This Physics by Emmy3(m): 12:04pm On Oct 15, 2012 |
S=(29x58) + [1/2(-1x58^2)] |
| Re: Help With This Physics by Emmy3(m): 12:07pm On Oct 15, 2012 |
S=1682-1682 =0m therefore, the train was brought to rest at the application of the brake by the Engineer. . |
| Re: Help With This Physics by SpicyMimi(f): 12:42am On Dec 30, 2012 |
Olodo! Na here u kon dey carry your expo abi? I am so ready for you...you ve just got yourself a pest!!! |
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