Maximizing Cyclist Speed: Solving the Energy Question in a Race

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A cyclist with an initial speed of 21 km/h and power output of 300 W can convert 92% of the input energy into kinetic energy over 4 seconds. The total energy input calculates to 1,200 J, leading to a useful energy of approximately 1,104 J. Using the kinetic energy formula, the change in velocity is determined to be +5.32 m/s, resulting in a final speed of 11.15 m/s. The discussion emphasizes the importance of calculating both initial and final kinetic energy to find the cyclist's new speed accurately. The participants clarify the calculations and formulas involved in deriving the final velocity.
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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.
 
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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 (3 S.F)
Total speed = 5.32+5.83 = 11.15m/s
 
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(Vfinal2-Vinitial2)/2
 
I don't understand your solution. Can you please Clarify
 
neededthings said:
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 (3 S.F)
Total speed = 5.32+5.83 = 11.15m/s

Whoa, back it up a bit. You've got the energy input correct. What's your initial kinetic energy? Add the energy input to that to get the final kinetic energy. Use that to get a final velocity.
 
Kishlay said:
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(Vfinal2-Vinitial2)/2

initial kinetic energy + useful energy given= final kinetic energy
 
hope this will make you clear
 
I get 21 which is not the correct? Please Help
 
  • #10
never mind i got it.

Thanks alot
 
  • #11
he he he ...
if you have really got it then it is fine, if not we are always to help you...!
 
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