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Brilliant_brown [7]
3 years ago
5

Jim began a 153-mile bicycle trip to build up stamina for a triathlon competition. Unfortunately, his bicycle chain broke, so he

finished the trip walking. The whole trip took 5 hours. If Jim walks at a rate of 4 miles per hour and rides at 42 miles per hour, find the amount of time he spent on the bicycle.
Physics
1 answer:
NARA [144]3 years ago
5 0

consider the motion when Jim travels by bicycle

D = total distance to be traveled for the trip = 153 mile

v = speed of riding = 42 mph

t = time taken to complete the trip by bicycle = ?

using the equation

t = D/ v

inserting the values

t = 153/42

t = 3.64  h

while walking and riding :

T = total time taken to complete the trip by walking and riding = 5 hours

t ' = time of walking = T - t = 5 - 3.64 = 1.36 hours

v' = speed of walking = 4 mph

d' = distance traveled by walking = ?

distance traveled by walking is given as

d' = v' t'

d' = 4 x 1.36

d' = 5.44 miles

d = distance traveled by bicycle = D - d' = 153 - 5.44 = 147.56 miles

v = speed of riding = 42 mph

t = time spent on the bicycle = ?

time spent on the bicycle is given as

t = d/v

t = 147.56/42

t = 3.51 hours




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\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

\texttt{ }

\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}

T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}

T : T_2 = \sqrt{L} : \sqrt{2L}

T : T_2 = 1 : \sqrt{2}

\boxed {T_2 = \sqrt{2}\ T}

\texttt{ }

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\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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