a = ∆v/t
the distance for each is s = vt + 1/2 at^2
A jogger accelerates from rest to 2.86 m/s in 3.38 s. A car accelerates from 22.9 to 35.0 m/s also in 3.38 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 3.38 s?
2 answers
a = change in v / change in time
a) (2.86 - 0) / 3.38 = 0.846 m/s^2 (pitiful, g = about 10)
b) (35.0 - 22.9) / 3.38 = 3.58 m/s^2
c) Jogger goes (1/2)a t^2 = (1/2)(0.846)(11.4) = 4.83 meters
car goes vt+(1/2)at^2 =22.9*3.38+(1/2)(3.58)(11.4) = 77.4+20.4= 97.8 meters
difference = 97.8 - 4.8 = 93 meters
a) (2.86 - 0) / 3.38 = 0.846 m/s^2 (pitiful, g = about 10)
b) (35.0 - 22.9) / 3.38 = 3.58 m/s^2
c) Jogger goes (1/2)a t^2 = (1/2)(0.846)(11.4) = 4.83 meters
car goes vt+(1/2)at^2 =22.9*3.38+(1/2)(3.58)(11.4) = 77.4+20.4= 97.8 meters
difference = 97.8 - 4.8 = 93 meters