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3 years ago

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An elite Tour de France cyclist can maintain an output power of 450 W during a sustained climb. At this output power, how long w

ould it take an 88 kg cyclist (including the mass of his bike) to climb the famed 1100-m-high Alpe d'Huez mountain stage?

1 answer:

jeka57 [31]3 years ago

3 0

Answer:

T = 35' 08"

Explanation:

By definition, the power is the rate of change of energy, with respect to time:

$P = \frac{\Delta E}{\Delta t} (1)$

ΔE, in this case, must be equal to the change in the gravitational potential energy during the climb (neglecting friction), as follows:

$\Delta E = m*g*h = 88 kg*9.8m/s2*1100 m = 948640 J (2)$

From (1) and (2), being P= 450 W, we can solve for Δt, as follows:

$\Delta t = \frac{\Delta E}{P} = \frac{948640J}{450W} = 2108 sec. (3)$

Converting seconds to minutes, we get:
Δt = 35' 08"

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$y_{1,o}$ - Initial height of the first stone, measured in meters.

$v_{1,o}$ - Initial speed of the first stone, measured in meters per second.

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$y_{2,o}$ - Initial height of the second stone, measured in meters.

$v_{2,o}$ - Initial speed of the second stone, measured in meters per second.

$t$ - Time, measured in seconds.

$t_{o}$ - Initial absolute time, measured in seconds.

$g$ - Gravity constant, measured in meters per square second.

Given that $y_{2,o} = 47\,m$ , $y_{2} = 0\,m$ , $t_{o} = 1\,s$ , $t = 2.866\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the following expression is constructed and the initial speed of the second stone is:

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$v_{1,o}, v_{1}$ - Initial and final velocities of the first stone, measured in meters per second.

$v_{2,o}, v_{2}$ - Initial and final velocities of the second stone, measured in meters per second.

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