Respuesta :
95 km/h = 26.39 m/s (95000m/3600 secs)
55 km/h = 15.28 m/s (55000m/3600 secs)
75 revolutions = 75 x 2pi = 471.23 radians
radius = 0.80/2 = 0.40m
v/r = omega (rad/s)
26.39/0.40 = 65.97 rad/s
15.28/0.40 = 38.20 rad/s
s/((vi + vf)/2) = t
471.23 /((65.97 + 38.20)/2) = 9.04 secs
(vf - vi)/t = a
(38.20 - 65.97)/9.04 = -3.0719
The angular acceleration of the tires = -3.0719 rad/s^2
Time is required for it to stop
(0 - 38.20)/ -3.0719 = 12.43 secs
How far does it go?
65.97 - 38.20 = 27.77 M
a) [tex]\rm \alpha =-3.0719\;rad/sec^2[/tex]
b) t = 12.43 sec
c) s = 95 m
Step by Step Solution :
Given :
Number of revolution (N) = 75
[tex]\rm v_1 = 95\times\dfrac{5}{18}=26.39\;m/sec[/tex]
[tex]\rm v_2= 55\times \dfrac{5}{18}=15.28\;m/sec[/tex]
Diameter = 0.80 m
Calculation :
[tex]\rm \omega = \dfrac{v}{r}[/tex]
[tex]\rm \omega_1 = \dfrac{26.39}{0.40}=65.94\;rad/sec[/tex]
[tex]\rm \omega_2 = \dfarc{15.28}{0.40}=38.20\;rad/sec[/tex]
We know that,
[tex]\rm t=\dfrac{2 \pi N}{\dfrac{\omega_1 + \omega_2}{2}}= \dfrac{2\pi\times 60}{\dfrac{65.94+38.20}{2}}[/tex]
[tex]\rm t = 9.04\;sec[/tex]
a) Angular Acceleration is ,
[tex]\alpha =\dfrac{\dfrac{\omega_2-\omega_1}{2}}{t}[/tex]
[tex]\alpha = \dfrac{\dfrac{38.20-65.24}{2}}{9.04}[/tex]
[tex]\rm \alpha =-3.0719\;rad/sec^2[/tex]
b) Time required by the car to stop
[tex]t = \dfrac{0-38.20}{-3.0719}[/tex]
[tex]\rm t = 12.43\;sec[/tex]
c) The distance covered by the car
[tex]\rm s = \dfrac{v^2-u^2}{2a}[/tex] --- (1)
[tex]\rm a = r \alpha[/tex]
[tex]\rm a = 0.40\times -3.0719=-1.2287\;m/sec^2[/tex]
From equation (1)
[tex]\rm s = \dfrac{0-15.28^2}{2\times-1.2287}=95\;m[/tex]
s = 95 m (Approx)
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