help me solve this: a bicyclist in the tour de france crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is 75 km/h. what was his average acceleration (in m/s square) while riding down the mountain?
First we have to convert: 75 km / h ( * 3.6 ) = 270 m/s 18 m/s = 64.8 m/s d = v i · t + 1/2 a · t² v = v i + a t ----------------------------------- 4000 = 64.8 · t + 1/2 a · t² 270 = 64.8 + a t a = ( 270 - 64.8 ) / t a = 205.2 / t 4000 = 64.8 t + 1/2 · ( 205.2 / t ) · t² 4000 = 64.8 t + 102.6 t 4000 = 167.4 t t = 4000 : 167.4 = 23.89 s a = 205.2 : 23.89 = 8.58 m/s² Answer: His average acceleration was 8.58 m/s²
