Calculate Cyclist's Speed in Race: Energy-Work Problem Solution

[neededthings]

Homework Statement

A cyclist is competing in a race and decides to pass some fellow cyclists who are getting tired. The initial speed of the cyclist is 21 km/h and he has 3.0 * 10^2 W of power. If the bicycle is able to convert 92% of input energy into kinetic energy, how fast will cyclist be traveling after 4.0 s. The combined mass of the cyclist and the bicycle is 78 kg.

The answer is 7.9 m/s. But I don't know how.

Homework Equations

p=e/t
Ek=1/2mv^2

The Attempt at a Solution

energy= Pt= 3x10^2 * 4 = 1.2 x10^3 J
Useful energy = 1.2 x10^3 *0.92 = 1.104 x10^3
KE= 1/2m( v)^2
root(1.104 x10^3 *2 /78) = change in v = +/-5.32 m/s (3 S.F.). He is accelerating hence +5.32m/s
21km/h = 21000m/60x60s= 5.83 m/s

 

[Kishlay]

why are you putting the same post every time?

 

[Kishlay]

go back to your previous post...!
i have posted solution over there

 

[Kishlay]

useful energy =1.104x10^3 ... right
then change in kinetic energy will be equal to useful energy given in 4 seconds...
the formula 1/2mv^2, 'V' is the instataneous velocity, not the change in velocity
so we get
change in kinetic energy = useful energy given..
mx(V_final²-V_initial²)/2

initial kinetic energy + useful energy given= final kinetic energy