Jimothy is riding his bike to his friend's house. He decides to see how fast he can go, and starts pedaling as hard as he can. He began with an initial velocity of 2.5 m/s. He pedals with an acceleration of 1.5 m/s^2 for a time of 6 seconds. Calculate his final velocity.

Final velocity, V = 11.5m/s

Explanation:

Given the following data;

Initial velocity, U = 2.5m/s

Acceleration, a = 1.5m/s²

Time, t = 6secs

To find the final velocity, we would use the first equation of motion

V = U + at

Substituting into the equation, we have

V = 2.5 + 1.5*6

V = 2.5 + 9

Final velocity, V = 11.5m/s

Explanation:

m₁ is heavy so it will go down and m₂ will go up. Let common acceleration be a .

For downward acceleration of m₁

m₁g - T₁ = m₁a

For upward acceleration of m₂

T₂- m₂g = m₂a

adding

m₁g - T₁ +T₂- m₂g = (m₁+m₂)a

m₁g- m₂g  +(T₂- T₁) = (m₁+m₂)a   ( 1 )

For rotatory motion of disc of mass mp and moment of inertia I

(T₁ -T₂ )R = I α , α is angular acceleration ( 2 )

From  1 and 2

m₁g- m₂g -  I α / R = (m₁+m₂)a

m₁g- m₂g -  I a / R² = (m₁+m₂)a

m₁g- m₂g = (m₁+m₂)a + I a / R²

m₁g- m₂g = [(m₁+m₂) + mp k² / R² ] a   ( k is radius of gyration of disk )

a = (m₁- m₂)g / [(m₁+m₂) +mp k² / R² ]

=  (m₁- m₂)g / [(m₁+m₂) + .5 mp  ]  ( for cylinder k² / R² = .5 )

If h be the height by which m₁ falls and its velocity becomes v

v² = 2 a h

a = v² / 2h

v² / 2h = (m₁- m₂)g / [(m₁+m₂) + .5 mp  ]

v²[(m₁+m₂) + .5 mp  ] = 2(m₁- m₂) gh

- v²[(m₁+m₂) +2(m₁- m₂) gh =  .5 mp v²

- v²[(m₁+m₂) +2(m₁- m₂) gh / .5 v ² = mp .

- 1.8² ( 5 + 3 ) + 2 ( 5 - 3 ) 9.8 x .75 / .5 x 1.8² = mp

- 25.92 + 29.4 / 1.62 = mp

2.15 kg = mp .

K.E  of masses = 1/2 x 5 x 1.8² + 1/2 x 3 x 1.8²

= 12.96 J

Rotational KE of disc

= 1/2 I ω²

= 1/4 mp r²ω²

= 1/4 x mp x v²

= .25 x 2.15 x 1.8²

= 1.74 J

in percent terms

(1.74 / 12.96) x 100

= 13.5 %