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Mathematics 9 Online
OpenStudy (anonymous):

Jose left the airport and traveled toward the mountains. Kayla left 2.1 hours later traveling 35 mph faster in an effort to catch up to him. After 1.2 hours Kayla finally caught up. Find Jose's average speed.

OpenStudy (anonymous):

lol.

OpenStudy (anonymous):

i duno dude this stuff doesn't make sense to me

OpenStudy (anonymous):

would kaylas rate be x+35 or 35 + x

OpenStudy (anonymous):

i thought the total time jose traveled was 3.3

OpenStudy (anonymous):

sorry got the names mixed up. Yeah Jose travelled for 2.1 + 1.2 hours which is 3.3 hours.

OpenStudy (anonymous):

and we know that Kayla travelled for 1.2 hours

OpenStudy (anonymous):

ok i got joses time = 3.3 hours and kaylas time = 1.2 hours and kaylas speed = x+35

OpenStudy (anonymous):

so i go d=(x+35)1.2

OpenStudy (anonymous):

yup and what does x signify?

OpenStudy (anonymous):

we don't need to know the distance...

OpenStudy (anonymous):

to find the distance then try to divide the distance by joses time to get his rate but its not the answer on the key

OpenStudy (anonymous):

The equation is Distance = Speed * Time The distance for which Kayla and Jose travel are the same. However, the speed and time at which they cover this distance is different.

OpenStudy (anonymous):

right

OpenStudy (anonymous):

Distance(Kayla) = Speed(Kayla) * Time(Kayla) Distance(Jose) = Speed(Jose) + Time(Jose)

OpenStudy (anonymous):

that why i found the distance with kayla first d=(x+35)1.2

OpenStudy (anonymous):

then used that distance and divided it by joses time

OpenStudy (anonymous):

because i see it as kayla is going faster then jose by 35mph so i go joses speed will be x plus 35 to get kaylas speed time her time 1.2

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

Distance(Kayla) = Distance(Jose) we can make it so, Speed(Kayla) * Time(Kayla) = Speed(Jose) * Time(Jose) So distance can be ruled out completely. We don't need to calculate distance at all.

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

but if we calculate distance can't we use it to find joses rate?

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

so would it be (x+35)1.2 = (x)3.3?

OpenStudy (anonymous):

Easier this way :P You got it... so multiply each side out.

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

i get 2x + 42 = 3.3x

OpenStudy (anonymous):

1.2x + 42 = 3.3x *

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

right my bad

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

then i minus the 1.2x from the 3.3x?

OpenStudy (anonymous):

so now move the x's to one side ... yeah take 1.2 x from each side

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

ok i got it now, dude what way are you doing it? ytou seem to be doing it differently then what im being taught

OpenStudy (anonymous):

to get the distance then take that distance and use it for joses time

OpenStudy (anonymous):

thank u though man

OpenStudy (anonymous):

As long as you're getting the right answer it's good. I just equate both sides... So i would have got the, 3.3x = 1.2x + 42 2.1 x = 42 x = 42/2.1 x is Joses speed. x = 20 and therefore x+35 is Kaylas speed which = 20 + 35 = 55

OpenStudy (anonymous):

Solve the following for Jose, jose's speed: (jose + 35) *1.2 = jose* 2.1 + jose 1.2 jose = 20 mph

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