As the car comes to rest it’s final velocity v will be 0
So our information g is
u = 12 m /s
t = 20s
v = 0
So acceleration is v-u/t
So 0-12/20
= -12/20
= -0.6 m / s^2
S= ut + 1/2 at^2
S = 12 x 20 + 1/2 x -12/20 x 400
S= 240 - 120
S = 120
It travels 120 m during the time interval.
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Mypages.valdosta.edu/bergstrm/audesrk21.htm
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
Twice as fast, so multiply the kinetic energy of the slower plane by two.
