Question
Coasting due west on your bicycle at 8.4 m/s, you encounter asandy patch of road 7.2m across. When you leave the sandy patch your speed hasbeen reduced by 2.0 m/s to 6.4m/s. Assuming the sand causes a constant acceleration, whatwas the bicycle's acceleration in the sandy patch? Give both magnitude and direction.How longdid it take to cross the sandy patch? Suppose you enter the sandy patch with speed of only 5.4 m/s.is your final speed in this case 3.4m/s, more than 3.4m/s, or lessthan 3.4m/s? Explain. Coasting due west on your bicycle at 8.4 m/s, you encounter asandy patch of road 7.2m across. When you leave the sandy patch your speed hasbeen reduced by 2.0 m/s to 6.4m/s. Assuming the sand causes a constant acceleration, whatwas the bicycle's acceleration in the sandy patch? Give both magnitude and direction.How longdid it take to cross the sandy patch? Suppose you enter the sandy patch with speed of only 5.4 m/s.is your final speed in this case 3.4m/s, more than 3.4m/s, or lessthan 3.4m/s? Explain.
Explanation / Answer
Vf^2 = Vo^2 +2ad 6.4^2 = 8.4^2 +2a(7.2) a= -2.1m/s^2 Theacceleration is due east at 2.1m/s^2 Vf = Vo +at 6 = 8 - 2.1t t=.95 ittook .95 seconds the final speed would have to be the same because accelerationis constant in the sand and the resistance should be the same sameno matter the velocity with which one goes through it