Problema Solution
A jogger accelerates from rest to 3.52 m/s in 2.52 s. A car accelerates from 21.2 to 39.4 m/s also in 2.52 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 2.52 s?
Answer provided by our tutors
(a)magnitude of acceleration of the jogger:
v=v_o+at
3.52=a(2.52)
a=1.40 m/s^2
(b)magnitude of acceleration of the car:
v=v_o+at
39.4=21.2+a(2.52)
a=7.22 m/s^2
distance of the jogger:
x=v_o(t)+(1/2)at^2
x=(1/2)(1.40)(2.52^2)
x=4.45 m
distance of the car:
x=v_o(t)+(1/2)at^2
x=21.2(2.52)+(1/2)(7.22)(2.52^2)
x=76.35 m
(c) how much further the car travels than the jogger:
distance=76.35-4.45
distance=71.9 m