Question

23) A mountain biker peddles his bike with cont rition from a zero in velocity standing start down asteep hill By the time he eaches the boo hill, he is traveling a velocity of 1.5 meters per second t o the b r28 to reach the bottom of the 4 ports each) CASNO ds (a) What was the biker's linear acceleration during this down ride? (Use Chapter 2) (b) How far did the biker travel during this downhill ride? (use Chapter 2) (c) If the diameter of the mountain bike tires - 68 meters, how many revolutions did the tues make during this downhill ride? (d) What was the angular velocity of the bike's tires when the biker reaches 18.5 meter per second? (e) What was the angular acceleration (a) of the mountain bike's tires during this downhill ride in radians per second squared? vo= 0 m/s V=115 m/s At = 285 v=votat => a: 175 m a S. 2 0 .675 m b) v²=v7 + 2a Ax => Ax v? ro - = 245 m za od 0-cu At = c) (=Q0r = 4.1 m 245 24. 60 revolutions all 0-0 = 1 (W + W)t 60 rev = 10.7985 - W = 4.3 rev - 27 rad 26.4 rad S. 285 0.95 rad = ST S-

help me solve this one pls

Answer 1

a) vf = 17.5 m/s
vi = 0 m/s
t = 28 s

the bikers linear acceleration, a = (vf -vi)/t

= (17.5 - 0 )/28

= 0.625 m/s^2 <<<<<<<<----------------Answer

b) distance travelled during the 28 s,

d = vi*t + (1/2)*a*t^2

= 0*28 + (1/2)*0.625*28^2

= 245 m <<<<<<<<----------------Answer

c) r = d/2 = 0.66/2 = 0.33 m

number of revolutions made, N = d/(2*pi*r)

= 245/(2*pi*0.33)

= 118 <<<<<<<<----------------Answer

d) use, w = v/r

= 18.5/0.33

= 56.1 rad/s <<<<<<<<----------------Answer

e) angular acceleration, alfa = a/r

= 0.625/0.33

= 1.89 rad/s^2 <<<<<<<<----------------Answer