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2 years ago

A cyclist rides in a circle with speed 5 m/s. What is his centripetal

Mathematics

1 answer:

Anna71 [15]2 years ago

6 0

Option A

The centripetal acceleration is 1.32 $ms^{-2}$

Solution:

From given question,

A cyclist rides in a circle with speed 5 m/s

The circle has a radius of 19 m

To find: Centripetal acceleration

The formula for Centripetal acceleration is given as:

$a_c = \frac{v^2}{r}$

Where,

$a_c$ = centripetal acceleration

r is the radius and v is the velocity

From given statement,

radius = r = 19 m

velocity = v = 5 m/s

Substituting the values in above formula, we get

$a_c = \frac{5^2}{19}\\\\a_c = \frac{25}{19}\\\\a_c = 1.315 \approx 1.32$

Thus centripetal acceleration is 1.32 $ms^{-2}$

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A. The city we will use is Orlando, Florida, and we are going to examine its population growth from 2000 to 2010. According to the census the population of Orlando was 192,157 in 2000 and 238,300 in 2010. To examine this population growth period, we will use the standard population growth equation $N_{t} =N _{0}e^{rt}$
where:
$N(t)$ is the population after $t$ years
$N_{0}$ is the initial population
$t$ is the time in years
$r$ is the growth rate in decimal form
$e$ is the Euler's constant
We now for our investigation that $N(t)=238300$ , $N_{0} =192157$ , and $t=10$ ; lets replace those values in our equation to find $r$ :
$238300=192157e^{10r}$
$e^{10r} = \frac{238300}{192157}$
$ln(e^{10r} )=ln( \frac{238300}{192157} )$
$r= \frac{ln( \frac{238300}{192157}) }{10}$
$r=0.022$
Now lets multiply $r$ by 100% to obtain our growth rate as a percentage:
$(0.022)(100)$ =2.2%
We just show that Orlando's population has been growing at a rate of 2.2% from 2000 to 2010. Its population increased from 192,157 to 238,300 in ten years.

B. Here we will examine the population decline of Detroit, Michigan over a period of ten years: 2000 to 2010.
Population in 2000: 951,307
Population in 2010: 713,777
We know from our investigation that $N(t)=713777$ , $N_{0} =951307$ , and $t=10$ . Just like before, lets replace those values into our equation to find $r$ :
$713777=951307e^{10r}$
$e^{10r} = \frac{713777}{951307}$
$ln(e^{10r} )=ln( \frac{713777}{951307} )$
$r= \frac{ln( \frac{713777}{951307}) }{10}$
$r=-0.029$
$(-0.029)(100)$ = -2.9%.
We just show that Detroit's population has been declining at a rate of 2.2% from 2000 to 2010. Its population increased from 192,157 to 238,300 in ten years.

C. Final equation from point A: $N(t)=192157e^{0.022t}$ .
Final equation from point B: $N(t)=951307e^{-0.029t}$
Similarities: Both have an initial population and use the same Euler's constant.
Differences: In the equation from point A the exponent is positive, which means that the function is growing; whereas, in equation from point B the exponent is negative, which means that the functions is decaying.

D. To find the year in which the population of Orlando will exceed the population of Detroit, we are going equate both equations $N(t)=192157e^{0.022t}$ and $N(t)=951307e^{-0.029t}$ and solve for t:
$192157e^{0.022t} =951307e^{-0.029t}$
$\frac{192157e^{0.022t} }{951307e^{-0.029t} } =1$
$e^{0.051t} = \frac{951307}{192157}$
$ln(e^{0.051t})=ln( \frac{951307}{192157})$
$t= \frac{ln( \frac{951307}{192157}) }{0.051}$
$t=31.36$
We can conclude that if Orlando's population keeps growing at the same rate and Detroit's keeps declining at the same rate, after 31.36 years in May of 2031 Orlando's population will surpass Detroit's population.

E. Since we know that the population of Detroit as 2000 is 951307, twice that population will be $2(951307)=1902614$ . Now we can rewrite our equation as: $N(t)=1902614e^{-0.029t}$ . The last thing we need to do is equate our Orlando's population growth equation with this new one and solve for t:
$192157e^{0.022t} =1902614e^{-0.029t}$
$\frac{192157e^{0.022t} }{1902614e^{-0.029t} } =1$
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$ln(e^{0.051t} )=ln( \frac{1902614}{192157} )$
$t= \frac{ln( \frac{1902614}{192157}) }{0.051}$
$t=44.95$
We can conclude that after 45 years in 2045 the population of Orlando will exceed twice the population of Detroit.

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Alla [95]

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