Respuesta :
Answer:
Explanation:
Given
initial velocity of motorcycle [tex]u=6.3\ m/s[/tex]
final velocity [tex]v=1.9\ m/s[/tex]
time taken [tex]t=4.6\ s[/tex]
radius of wheel [tex]r=0.68\ m[/tex]
using v=u+at
where a=acceleration
[tex]1.9=6.3+a\times 4.6[/tex]
[tex]a=-0.956\ m/s^2[/tex]
angular acceleration [tex]\alpha =\frac{a}{r}[/tex]
[tex]\alpha =\frac{-0.956}{0.68}=-1.406\ rad/s^2[/tex]
[tex]Displacement=average\ velocity\times time[/tex]
[tex]Displacement=\frac{u+v}{2}\times t[/tex]
[tex]Displacement=\frac{1.9+6.3}{2}\times 4.6=18.86\ m[/tex]
Angular displacement [tex]\theta =\frac{Linear\ displacement}{radius}[/tex]
Angular displacement [tex]\theta =\frac{18.86}{0.68}=27.73\ rad[/tex]
Answer:
Explanation:
initial velocity, u = 6.3 m/s
final speed, v = 1.9 m/s
time, t = 4.6 s
radius, r = 0.68 m
(a)
initial angular velocity, ωo = u / r = 6.3 / 0.68 = 9.26 rad/s
final angular velocity, ω = v / r = 1.9 / 0.68 = 2.79 rad/s
Let α be the angular acceleration
Use first equation of motion
ω = ωo + αt
2.79 = 9.26 + α x 4.6
α = - 1.41 rad/s²
(b) Let Ф be the angular displacement
Use third equation of motion
ω² = ωo² + 2 αФ
2.79² = 9.26² - 2 x 1.41 x Ф
Ф = 27.6 rad
