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nignag [31]
2 years ago
11

A motorcycle starts from rest and maintains a constant acceleration whose magnitude is 0.45 m/s².

Mathematics
1 answer:
natali 33 [55]2 years ago
3 0

a) If the motorcycle starts from rest, then we have an initial speed equal to zero (V0 = 0), with this we can choose a formula that does not contain the initial speed, now we also know that what it asks us is the time that it takes a certain speed to pass, so the formula we will use will be:

\bf{\displaystyle d=\frac{a{{t}^{2}}}{2}}

Clearing "t"

\boldsymbol{\sf{t=\sqrt{\dfrac{2d}{a}  }  } }

The distance is in Kilometers, we first need to pass this distance in meters, so we'll need to do our conversion:

\boldsymbol{\sf{1.6\not{km}*\dfrac{1000 \ m}{1\not{km}}=1600 \ m  } }

1.6 km = 1600m

Substituting our data into the formula:

\boldsymbol{\sf{t=\sqrt{\dfrac{2d}{a}  }=\sqrt{\dfrac{2(1600 \ m) }{9.45 \ \frac{m}{s^2}  }  }=84.32 \ s   } }

So the motorcycle will take 84.32 seconds to travel 1600 meters.

b) Since we are asked to obtain the speed in the time it took for the 1600 meters, we are going to use the following formula.

\boldsymbol{\sf{V_{f}=at }}

Substituting our data:

\boldsymbol{\sf{V_{f}=at=\left(0.45 \ \dfrac{m}{\not{s^{2}} }\right)(84.23 \ s)=37.94 \ \dfrac{m}{s}   }}

In order to express the final result of the speed, in terms of km/h, we would only apply the conversion factors.

\boldsymbol{\sf{V_{f}=37.94\dfrac{\not{m}}{\not{s}}*\dfrac{1 \ km}{1000\not{m}}*\dfrac{3600\not{s}}{1 \ h}=136.58 \ \dfrac{km}{h}     }}

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652.6-2.01\frac{311.7}{\sqrt{50}}=563.997    

652.6+2.01\frac{311.7}{\sqrt{50}}=741.203    

So on this case the 95% confidence interval would be given by (563.997;741.203)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=652.6 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=311.7 represent the sample standard deviation

n=50 represent the sample size  

Soltuion to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=50-1=49

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that t_{\alpha/2}=2.01

Now we have everything in order to replace into formula (1):

652.6-2.01\frac{311.7}{\sqrt{50}}=563.997    

652.6+2.01\frac{311.7}{\sqrt{50}}=741.203    

So on this case the 95% confidence interval would be given by (563.997;741.203)    

6 0
3 years ago
Daniel is currently 26 years older than his son. In sic years he will be tree times older than his son. How old are both of them
Vlad1618 [11]

well, 26+6 is 32, so divide that by 3 and you'd get 10.67, making daniel 32 by that time and his son being 10 years old

its simple math really, also, i hope this helps you ^-^

4 0
4 years ago
Read 2 more answers
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