Since the acceleration is constant, the average velocity is simply the average of the initial and final velocities of the body:
We can proof that the distance covered by the body moving at constant average velocity 
is equal to the distance covered by the body moving at constant acceleration a:
- body moving at constant velocity : distance is given by
- body moving at constant acceleration 
: distance is given by
Answer:
Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.
incline angle = θ ; can's mass = m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2
T1 = Inertia of Can with high viscosity soup
T2 = Inertia of Can with low viscosity soup
M1 rolling moment of Can 1
M2 rolling moment of Can 2
equation is given by
T1*α1 = M1 - (a)
T2*α2 = M2 - (b)
M1 = M2 = m*g*R*sin(θ). (c)
as assumed T1 > T2
from the three equation (a), (b) & (c)
the α2 > α1
Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.
Answer: final Velocity v = 10.2m/s
Explanation:
Final speed v(t) is given as
v(t) = u + at .......1
Where; u = the initial speed
a = acceleration
t = time taken
The total distance travelled d is given as
d = ut + 1/2(at^2)
Given
d = 5.0m
u = 2.0m
a = g = 10m/s2 (acceleration due to gravity)
Substituting into the equation above we have
5 = 2t + 5t^2
5t^2 +2t -5 = 0
Applying the quadratic formula. We have;
t = 0.82s & t = -1.22s
t cannot be negative
t = 0.82s
From equation 1 above
v = 2.0m/s + 10(0.82)m/s
v = 10.2m/s
Answer:
a = 2.72 [m/s2]
Explanation:
To solve this problem we must use the following kinematics equation:
where:
Vf = final velocity = 1200 [km/h]
Vo = initial velocity = 25 [km/h]
t = time = 2 [min] = 2/60 = 0.0333 [h]
1200 = 25 + (a*0.0333)
a = 35250.35 [km/h2]
if we convert these units to units of meters per second squared
![35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]](https://tex.z-dn.net/?f=35250.35%5B%5Cfrac%7Bkm%7D%7Bh%5E%7B2%7D%20%7D%5D%2A%28%5Cfrac%7B1%7D%7B3600%5E%7B2%7D%20%7D%29%2A%5B%5Cfrac%7Bh%5E%7B2%7D%20%7D%7Bs%5E%7B2%7D%20%7D%20%5D%2A%28%5Cfrac%7B1000%7D%7B1%7D%20%29%2A%5B%5Cfrac%7Bm%7D%7Bkm%7D%20%5D%20%3D%202.72%20%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D)
