A 2,00kg car slowed down uniformly from 20.0m/s to 5.00 m/s in 4.00 s.
(a) what is the average force acted on the car during that time and (b) how far did the car travel during that time?
2007-06-05 05:29:26 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
From the second law of Newton you will find the force. The equation is F=m*a you have the m so you have to find the a (acceleration)
U=Uo-a*t (5=20-a*4) so a=15/4
F=m*a so F = 2*15/4=7,5N
X=Uot-1/2a*t^2 so x = 20*4 - 1/2 *15/4*16 = 50m
2007-06-05 05:39:15 · answer #1 · answered by Anonymous · 0⤊ 0⤋
u = 20 m/s and v = 5 m/s and t = 4 secs
v = u + at is the formula to be used.
5 = 20 + a.4 or 4a = -15 and a = -15/4 m/s/s
F = m.a = 200 x -15/4 = -750 N (I feel the mass should be 2000kg and not 200 kg). If m is 2000 kg F = -7500 N (indicating that it is a braking force)
s = ut + 1/2 a t^2 = 20X4 + 1/2 X -15/4 X 4 X 4
= 80 - 30 = 50m
2007-06-05 12:38:34 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋