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2 years ago

A swimmer speeds up 1.9 m/s to 3.8 m/s during the last 11.0 seconds of the race. What is the acceleration of the swimmer?

Mathematics

1 answer:

Naddika [18.5K]2 years ago

8 0

<h2>Given :-</h2>

Initial speed (v) = 1.9m/s

Final speed (u) = 3.8m/s

Time (t) = 11 s

<h2>Formula Used :-</h2>

$v = u + a×t$ (a = acceleration)

➾ $a = {\frac{u-v}{t}}$

<h2>Solution:-</h2>

$a = {\frac{u-v}{t}}$ (putting the value of v, u and t from above)

➾ $a = {\frac{3.8-1.9}{11}}$

➾ $a = {\frac{1.9}{11}}$

➾ $a = 0.17 m/s (approx.)$

Therefore, the acceleration rate = 0.17m/s.

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$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

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Step-by-step explanation:

Data given and notation

$\bar X_{1}=9.1$ represent the mean for the sample 1

$\bar X_{2}=8$ represent the mean for the sample 2

$s_{1}=1.9$ represent the sample standard deviation for the sample 1

$s_{2}=2.1$ represent the sample standard deviation for the sample 2

$n_{1}=40$ sample size for the group 1

$n_{2}=50$ sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

Statistical decision

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=40+50-2=88$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(88)}>2.604)=0.0108$

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