Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
The initial kinetic energy of the car is
Then, the velocity of the car is decreased by half:
so, the new kinetic energy is
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
Is there a multiple choice?
I am going to assume 2.1 metres per second and that we're rounding acceleration due to gravity to -10 metres per second squared. At the highest point, velocity is going to be 0. v= intial velocity + acceleration*time, sub in 0 for velocity, 2.1 for initial velocity and -10 for acceleration to get 0= 2.1-10t. Now solve for t. t=0.21 seconds.
