A race car starts from rest on a circular track of radius 400 m. Its speed increases at the constant rate of 0.5 m/s. At the point where the magnitudes of the radial and tangential accelerations are equal, determine (a) the speed of the race car, and (b) distance traveled, (c) the elapsed time.
2006-11-07 14:49:13 · 4 answers · asked by thefishbag 2 in Science & Mathematics ➔ Physics
i think you made a mistake its supposed to be 0.5m/s^2.
then your answers are
a) a=v^2/r
0.5=v^2/400
200=v^2
v= 14.14m/s
b) v^2=u^2+2as
200=0+(2*0.5*s)
s=200m
c) s= ut + 0.5 at^2
200 = 0 + (0.5* 0.5*t^2)
800 =t^2
t= 28.3s
2006-11-07 18:45:33 · answer #1 · answered by gerry zim 2 · 0⤊ 0⤋
around action deals with in basic terms the action which could be in around yet in rotational motions it would desire to not be in around even with the undeniable fact that it would be in rotational with admire to any of its appropriate factor.
2016-11-28 02:48:21 · answer #2 · answered by ? 4 · 0⤊ 0⤋
tangential acceleration: 0.5m/s^2 (given)
radial acceleration :v^2/r
equate the two
a)v=sqrt(400*0.5)=14.14m/s
b)d=a*t^2, but a=(v_final-v_i)/t=v/t, t=v/a
d=a*(v/a)^2=v^2/a=R=400m
c)t=v/a=14.14/0.5=28.284s
2006-11-07 17:56:09 · answer #3 · answered by justiceforall 2 · 0⤊ 0⤋
Use centripetal acceleration equation to find the speed.
2006-11-07 17:43:53 · answer #4 · answered by arbiter007 6 · 0⤊ 0⤋