Question

I promise I will mark as brainiest if a speed of a car increase its average speed in a journey of 200 miles by 5 mile/hour the journey will end with a difference of one hour less, so what is the original speed of the car ?

Answer 1

Answer:

$\large \boxed{\sf \ \ \ \dfrac{\sqrt{4025}-5}{2}=29.22144... \ \ \ }$

Step-by-step explanation:

Hello

Let's note the original speed of the car v

it means that in 1 hour he is going v miles

so to go 200 miles it takes ( in hour)

$\dfrac{200}{v}$

If the speed of the car is v+5 than to go 200 miles it takes (in hour)

$\dfrac{200}{v+5}$

and this time is one hour less so we can write

$\boxed{\sf \ \ \dfrac{200}{v+5}=\dfrac{200}{v}-1 \ \ }$

We can multiply by v(v+5) both parts of the equation so

$200v=200(v+5)-v(v+5)\\\\200v=200v+1000-v^2-5v\\\\v^2+5v-1000=0$

$\Delta=b^2-4ac=5^2+4*1000=4025\\\\ \text{There are potential solutions }\\\\\ \ \ \ \ x_1=\dfrac{-5-\sqrt{4025}}{2}\\\\\ \ \ \ \ x_2=\dfrac{-5+\sqrt{4025}}{2}$

Only one is positive and this is is

$x_1=\dfrac{\sqrt{4025}-5}{2}=29.22144...$

So the original speed is 29.22144... mph

Hope this helps.

Do not hesitate if you need further explanation.

Thank you