motorcycles brakes are designed to absorb 35kJ/s, the motorcycle and rider have a mass of 280kg and are traveling at 100km/h.how fast will they be going after 3 seconds of braking?thanks
V1 100 Km/h 100000 m / 3600 s 27.78 m/s the absorbed energy after 3 s 35000 x 3 105000 J so 105000 KE1 - KE2 1/2 m (V1^2 - V2^2) 105000 27.78^2 - V2^2 105000 / 140 750 V2^2 21.73 hence V2 4.66 m/s
I need a rubber stamp. Please use m/s not ft/sec, miles/hr, km/hr. The reason is that all physics formulae are defined in terms of SI units. 100 Km /hr is just a rounded approximate number. So it would have been just as easy to use a rounded number in m/s ( 30 is pretty close) But to use the exact figures given. V 100 * 1000 / 3600 27.8 m/s (approx) E 1/2 m v^2 1/2 * 280 * (100 * 1000 / 3600) ^ 2 1.08 * 10 ^ 5 J In 3 secs the brakes will remove 35 * 10 ^ 3 * 3 1.05 * 10^5 J Subtract this from the energy available. now to get the speed v sqrt( 2 E / m) sqrt( 2 * ( 1.08 - 1.05)* 10 ^ 5 / 280) Now given that the 35 is only two significant figures and 3 sec is only one significant figure then the answer is close to zero You would only get a valid answer if you presumed that 100 Km/hr meant 100.0000Km/hr and that 35 KJ/s meant 35.0000 KJ/s and that time meant 3.00000s Arithmetically this gives 4.6481113 m/s but this is not scientifically valid. Only mathematically.
