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2 years ago

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Your friend states in a report that the average time required to circle a 1.5-mi track was 65.414 s. This was measured by timing

7 laps using a clock with a precision of 0.1 s. How much confidence do you have in the results of the report? Explain.

1 answer:

Aleks04 [339]2 years ago

5 0

The time per lap was calculated by measuring the time for seven laps and dividing the total time by seven.

total time $65.414 s \times 7 =457.898 s.$

It is given that the precision is of 0.1 s. it means it is correct upto 1 place beyond decimal.

So, the actual value could vary from 457.800 s to 457.899 s i.e. the time per lap could be 65.400 s to 65.414 s

It means measured time 65.414 s has maximum error of 0.021%. Hence, the measured value is quite precise.

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Answer:

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(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

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Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

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Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

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$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

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