A car travels 65 km in 45 minutes what is the average speed of the car in km/h

2 answers:

5 0

So to begin your problem, you know that your car already has an average which is 65km/45 mins. The problem wants you to change this to km/hr. This means that you need to convert minutes to hours. A simple way to do this is by using fractions.
Set your problem up with fractions similar to this:
65km/45 mins x 60 mins/1 hr.
the whole point is to cancel out your minutes, and leave the hours as your new unit for the denominator
65km/45 x 60/1 hr.
now you want to reduce (I divided the first fraction by 5)
13km/9 x 60/1 hr.
780km/9hrs.
That would be your answer. If someone can double check my math that would be fantastic.

mestny [16]3 years ago

3 0

65km in 45 min
1 hour=1h=60min
45/60=3/4
65km=3/4hour
multiply both sides by 4/3 to clear fraction
260/3=1 hour
86 2/3 =1 hour
aprox
86.666666666 km per hour

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Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$