Question

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2 h, and Car B traveled the distance in 1.5 h. Car B traveled 15 mph faster than Car A.



How fast did Car B travel?

Answer 1

@StudyGurl14

Answer 2

D = distance, S = speed, T = time, A = car A, B = car b

\[D_A=D_B\]
\[S_A+15=S_B\]
\[T_A =2\]
\[T_B=1.5\]
And remember...
\[S = \frac{ D }{ T }\]
Can you do the rest?

Answer 3

@StudyGurl14 How do you get speed A? All I know is it was 15 mph slower than car B

Answer 4

\[S_A=\frac{ D_A }{ T_A} ; S_B = \frac{D_B}{T_B}\]
Does that help?

Answer 5

So the distance divided by time is the speed of car A, but i dont know the distance.

Answer 6

@StudyGurl14

Answer 7

Because they travel the same distance, you can just use D instead of D_A and D_B...
\[S_A=\frac{D}{T_A}\]
\[S_B = \frac{D}{T_B}\]
\[D=S_A \times T_A ; D = S_B \times T_B\]
That should help.

Answer 8

I'm still very confused about this question. I don't know speed or distance. all I know is time and that B was 15 mph faster. How do you divide a letter?

Answer 9

@StudyGurl14 Is it 60?

Answer 10

YES! OMG great job!

Answer 11

do you know how to do it, or did you find that on the internet? do you want me to show you the right way to do it?

Answer 12

@StudyGurl14 I failed the test, and it said that was the correct answer. I thought it was 50

Answer 13

i got 2 questions wrong on the test :S

Answer 14

@StudyGurl14 Can you show me the right way to do it? I need to know to redo the test.

Answer 15

two questions wrong isn't failing...

Answer 16

when there are 5 questions it is.

Answer 17

oh...then it is a quick check. who cares. those are only worht 5% of grade

Answer 18

do you want me to show you how?

Answer 19

it was an assesment, yes please

Answer 20

okay. hold on.

Answer 21

You have all this information:
\[S_A=\frac{D_A}{T_A} ; S_B=\frac{D_B}{T_B}\]
\[D_A=S_A \times T_A ; D_B=S_B \times T_B\]
\[D_A=D_B\]
so...
\[D = S_A \times T_A ; D=S_B \times T_B\]
so...
\[S_A \times T_A = S_B \times T_B\]
And we know...
\[S_A+15=S_B\]
\[T_A = 2 ; T_B = 1.5\]
Rearrange:
\[S_A = S_B - 15\]

Substitute:
\[(S_B-15) \times (2) = S_B \times (1.5)\]
Solve for S_B (which is the speed of car B)

Can you do the rest yourself?

Answer 22

@StudyGurl14 it is like you are speaking a foreign language. I don't understand it.

Answer 23

What don't you understand? Be specific

Answer 24

How you got
(SB−15)×(2)=SB×(1.5)

Answer 25

or how to solve

Answer 26

(Sb-15)x(2)=Sbx(1.5)*

Answer 27

Okay, let me finish it for you....
\[(S_B-15)\times (2) = S_B \times (1.5)\]
\[2S_B - 30 = 1.5S_B\]
\[-30 = -0.5S_B\]
\[S_B = 60\]

Answer 28

still confused?

Answer 29

Ohhhh

Answer 30

I have this one, if you want me to open up another question so you get another medal i will Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 4 h, and Car B traveled the distance in 4.5 h. Car B traveled 5 mph slower than Car A.



How fast did Car B travel?

Answer 31

How do you get the equation?

Answer 32

open new question please

Answer 33

how did i get what equation?

Answer 34

@25Mattman

Answer 35

never mind, sorry @StudyGurl14

Answer 36

lol ok