Question

Rachel runs 2km to her bus stop, and then rides 4.5 km to school. On average, the bus is 45 km/h faster than Rachel's average running speed. If the entire trip takes 25 minutes, how fast does Rachel run?

Answer 1

Answer:6.042km/h 6km/h approximately

Step-by-step explanation:

First off we have to know the formula relating speed, distance and time which is

Speed = distance/time

Now we are looking for Rachel's running speed

We are to find Rachel's running speed, so let's label is x

We are given that the distance Rachel runs to her bus stop is 2km

We were not given the time she uses to run to the bus stop

So let's label the time Rachel uses to run to her bus stop as y

So from the formula speed = distance/time

We have x = 2/y

Now we are told that the speed the bus uses to get to school is 45km/h faster than her speed used to run

So speed of bus = 45 + x

And the overall time for the whole journey is 25mins, changing this to hours, because the speed details given is in km/h we divide 25 by 60 which will give 0.417

Now if the total time is 0.417 hours, and we labeled the time for Rachel to run to the bus as y, so the time for the time for the bus to get to school will be 0.417 - y

We are also told the bus rides for 4.5km to school

So adding this together to relate the speed, distance and time of the bus with the formula speed = distance/time

We get 45 + x = 4.5/(0.417 - y)

So we have two equations

x = 2/y (1)

45+x = 4.5/(0.417-y) (2)

So putting (1) in (2) we have

45 + (2/y) = 4.5/(0.417-y)

Expanding further

(45y + 2)/y = 4.5(0.417-y)

Cross multiplying

(45y + 2)(0.417 - y) = 4.5y

Opening the brackets

18.765y - 45y2 + 0.834 - 2y = 4.5y

Collecting like terms

-45y2 + 18.765y -2y - 4.5y + 0.834 = 0

-45y2 + 12.265 + 0.834 = 0

Dividing all sides by -45 to make the coefficient of y2 1

y2 - 0.273y - 0.019 = 0

Now we have gotten a quadratic equation, and since it's with decimal numbers we can use either completing the square method of almighty formula

I'm using almighty formula her

For solving

ax2 + bx + c = 0

x = (-b +-root(b2-4ac)/2a

For our own equation, we are finding y

From our our quadratic equation

a = 1, b=-0.273, c = -0.834

you = (-(-0.273)+-root(-0.273-4(1)(-0.019))/2(1)

y = (273+-root(0.151))/2

y = (0.273+0.389)/2 or (0.273-0.389)/2

y = 0.331 or -0.085

So we use the positive answer which is 0.331, because time can't be negative

Then we put y = 0.331 in (1)

x = 2/y

x = 2/0.331

x = 6.042km/h

x = 6km/h approximately