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IrinaVladis [17]

3 years ago

15

A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bic

yclist’s acceleration to the nearest tenth of a m/s^2 ?

1 answer:

zalisa [80]3 years ago

3 0

Answer:

The bicyclist's acceleration is 2.2m/s^2

Explanation:

Given

$u = 2.5m/s$ ---- Initial Velocity

$v = 12.5m/s$ ---- Final Velocity

$t = 4.5s$ ---- Time

Required

Determine the acceleration

This will be solved using the first equation of motion

$v = u + at$

Substitute values for v, u and a

$12.5 = 2.5 + a * 4.5$

$12.5 = 2.5 + 4.5a$

Collect Like Terms

$4.5a = 12.5 - 2.5$

$4.5a = 10.0$

Solve for a

$a = 10.0/4.5$

$a = 2.2m/s^2$ ---- (approximated)

<em>Hence, the bicyclist's acceleration is 2.2m/s^2</em>

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

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Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

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Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

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Isolating $m$ :

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<h3>b) Period</h3>

The period $T$ is given by:

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Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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