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stealth61 [152]

3 years ago

12

A car passes a landmark on a highway traveling at a constant rate of 45 kilometers per hour. One hour later, a second car passes

the same landmark traveling in the same direction at 65 kilometers per hour. How much time after the second car passes the landmark will it overtake the first car?

2 answers:

katrin2010 [14]3 years ago

5 0

<u>Answer:</u>

2 hours 15 mins

<u>Step-by-step explanation:</u>

We know,

the speed of car 1 = 45 kilometers per hour; and

the speed of car 2 = 65 kilometers per hour

Assuming the time for the 2nd car to catch the 1st one to be t, we can write the distance equation:

$65t=45t+45$

$65t-45t=45$

$20t=45$

$t=45/20$

$t=2.25$

2.25 hours = 2 hours + (0.25 * 60) min = 2 hours 15 mins

Therefore, it takes 2 hours 15 mins for the second car to overtake first car after it passes the landmark.

vazorg [7]3 years ago

3 0

A car passes a landmark on a highway traveling at a constant rate of 45 kilometers per hour.

Let t be the time taken by second car

So t+1 is the time taken by first car

Distance = speed * time

Distance traveled by first car = 45 * (t+1)

second car passes the same landmark traveling in the same direction at 65 kilometers per hour

Distance traveled by second car = 65 * (t)

When second car overtakes the first car then their distance are same

65 t = 45(t+1)

65t = 45t + 45

Subtract 45 t from both sides

20t = 45

Divide both sides by 20

so $t = \frac{45}{20}=\frac{9}{4}=2.25$

It took 2.25 hours for the second car to overtake first car

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b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

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We can apply natural log in both sides and we got:

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If we multiply by -1 both sides of the inequality we have:

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Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

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The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

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$t \geq \frac{-ln(0.2)}{2}$

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And then we can conclude that the time period with any load would be 0.8047 years.

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