Answer:
0.108 rad/s².
Explanation:
Given that
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 0.5 rev/s
We know that
1 rev/s = 6.28 rad/s
ωf= 3.14 rad/s
t= 28.9 s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
3.14 = 0 + α x 28.9
Therefore the acceleration will be 0.108 rad/s².
Therefore the answer will be 0.108 rad/s².
Answer:
<em>The cyclist is traveling at 130 m/s</em>
Explanation:
<u>Constant Acceleration Motion
</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

The cyclist initially travels at 10 /s and it's accelerating at a=6m/s^2. We need to know the new speed when t= 20 seconds have passed.
Apply the above equation:



The cyclist is traveling at 130 m/s
Zero acceleration means the object's velocity is not changing.
So the object is moving in a straight line, at a constant speed
that could be anything (including zero) as long as it's constant.
Answer:
a.3.84m
b.-106.67m/s
c.947.3m/s^2
d.70.17 rad
e.2.5Hz
d.0.4secs
Explanation:
Given x=(7.8)cos[5πrad/s)t+π/3)]
a.Displacement at t=4.4
7.8cos(5π*4.4+π\3)=3.84m
b.velocity
V= dx/dr=-5π(7.8)sin(5πrad/s)t+π\3
at t=4.4
-5π(7.8)sin(5π*4.4+π\3)=-106.67m/s
c.acceleration
a=d^2x/dr^2
-(5π)^2(7.8) cos (5π*t+π\3)
at t=4.4
-(5π)^2(7.8)cos(5π*4.4+π\3)=-947.3m/s^2
d. Phase =(5πrad/s)t+π\3
At t=4.4
5π×4.4+π\3=70.17 rad
e.frequency
Given x= 7.8cos(5πt+π\3
Compare with x=Acos(2πft)
2πft=5πt
F=2.5Hz
f.T=1\f
T=1/2.5=0.4sec
Answer:
it's a little hard I may help you after