please help! on math question! its suppose to be an easy question for most people, but not easy for me!?

Question

please dont just give me the answer, i really do want to learn, please explain step by step, because im really bad at math.

Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00am, each heading for Wildwood. One car's average speed is 10 miles per hour more than the other's. The faster car arrives at Wildwood at 11:00 am, 1/2 hour before the other car. What is the average speed of each car? How far did each travel?

Answer 1

Car 1""""""""""Car 2
x mph"""""""""x + 10 mph
3.5 h"""""""""""""3 h
d = 3.5 x""""""3(x + 10)

3.5x = 3x + 30
0.5 x = 30
x = 60
Speed are 60 mph and 70 mph
d = 3.5 x 60 = 210 miles

Answer 2

Remember distance=rate * time or d-rt
Since we are dealing with mph (miles per hour) our unit of time must be expressed in hours and our unit of distance must be expressed in miles.

Let x equal the average speed of car 1
We can now assume that the average speed of car 2 is x+10

If car 2 (the faster car) arrives at 11:00 and the problem tells you that it arrives "1/2 hour before the other car", we can conclude that car 1 must have arrived at 11:30
Therefore, car 2 took 3 hours to travel (8am to 11am) and car 1 took 3 1/2 hours.

So now we have the rate and time, so we can write our equations using the d=rt formula:

d(car2)=3x
d(car1)=3 1/2 (x+10)

It will make things a lot easier if we can rework that second equation by simplifying.

d=(7/2 )x+(7/2)(10)
d=(7/2)x+(70/2)
d=(7/2)x+35


Which is farther, the distance from Commercial Boulevard to Wildwood or the distance from Commercial Boulevard to Wildwood? Both cars took the exact same route and therefore traveled the exact same distance. Therefore, we know that the distance traveled by car 1 is equal to the distance traveled by car 2. Because of this, we can set our two equations equal to one another as such:

(7/2)x+35=3x
multiplying by 2 will get rid of that 7/2 and make things easier:
7x+70=6x
factor out a 7
7(x+10)=6x
divide both sides by 7
x+10=(6/7)x
10=(-1/7)x
-70=x
and x+10 would be -60

Since you are dealing with average speed, don't worry about the negative signs. One car traveled 60 mph while the other one traveled 70mph

For distance, you would just need to multiply by 3 1/2 and 3 respectively.

Answer 3

OK, Both cars went the same distance. The faster car got there in 3 hours, and the slower car got there in 3.5.

If we say A is the speed of the faster car, and B is the speed of the slower car, we know two major things about them:

A = B + 10

since they both went the same distance( = hours x average speed), then:

3 * A = 3.5 * B

so if we substitute (B+10) for A in the second equation, we get:

3 * (B + 10) = 3.5 * B

which rapidly reduces to

30 = 1/2 B

or B = 60

this would mean A=70. If A got there in 3 hours with an average speed of 70, than the distance was 210 miles. As a check we multiply 3.5 hours by 60 miles per hour, and we also get 210 miles, which we should since they went that same distance.

Answer 4

let speed of slower car = x mi/hr
the faster car speed = x + 10 mi/hr

the time taken by faster car (11am - 8 am) = 3 hrs
the time taken by slower car = 3 +1/2 hrs = 7/2 hrs

distance travlled by faster car = speed * time = 3(x + 10)
distance travelled by slower car = 7/2 x

3(x + 10) = 7/2 x ( since both travelled same distance)

multiply with 2
6(x + 10) = 7x
6x + 60 = 7x

slower car speed =x = 60 mi/hr
faster car speed = x + 10 = 70 mi/hr
each car travelled = 70*3 or 60*(7/2) = 210 miles

Answer 5

first you need to determine what you know.
a) you know the distance each car travelled is the same (this will be important later)
b) car1 has a speed that is 10mph greater than car2
c)car1's time is 3 hrs
d)car2's time is 3.5 hrs
e) you also know the equation d=r*t
r1 = car1's speed
r2 = car2's speed
d1 = distance travelled by car1
d2 = distance travelled by car2
t1 = time travelled by car1
t2 = time travelled by car2

now, you need to define your unknown. if you say that r1 = x; you can say that r2 = (x - 10) b/c car 2 is slower than car 1. if that confuses you, define it the other way, r2 = x, then r1 = (x + 10) (b/c car1 is faster.) since you say what x is, you can do this.

here is where the distance being the same comes into play. since they travel an equal distance, you can set their distance equations equal to each other, so:
r1*t1=r2*t2

now, just plug your values in and solve for x.

remember to refer back to your definition of x when you get your number.

x will be the speed of the car that you defined it as. you will need to either add or subtract 10 (depending on which definition you used) to get the other speed.

Answer 6

distance = speed times time. Both traveled the same distance, so speed times time for both must also be the same. If the faster car's speed was s, the slower was s - 10. The faster time was 3 hours and slower was 3.5 hours

So, s times 3 = (s-10) times 3.5

3s = 3.5(s-10)

do it out by distributive property

3s = 3.5s - 35

add -3.5s to both sides

-0.5s = -35

divide both sides by -0.5 and you get the faster speed. Use that to find the slower speed. Then multiply the faster speed by 3 because distance = speed times time. to get the distance.

Answer 7

distance = speed times time. Both traveled the same distance, so speed times time for both must also be the same. If the faster car's speed was s, the slower was s - 10. The faster time was 3 hours and slower was 3.5 hours

So, s times 3 = (s-10) times 3.5

3s = 3.5(s-10)

do it out by distributive property

3s = 3.5s - 35

add -3.5s to both sides

-0.5s = -35

divide both sides by -0.5 and you get the faster speed. Use that to find the slower speed. Then multiply the faster speed by 3 because distance = speed times time. to get the distance. Good luck !!! Hope this Helps !!! =D

Answer 8

d=distance to Wildwood, v is slower car's speed.

d/(v+10)=3hr d/v=3.5hr =>d=3.5v
3.5v=3v+30 =>v=60 => d=3.5v=210mi.
average speeds are 60 and 70 and each drove 210 miles.

Answer 9

The first car traveled for 3 hours and went 3x miles, where x is its speed in miles per hour.

The second car traveled for 3.5 hours and went 3.5(x-10) miles since (x-10) is its speed in miles per hour.

The two cars went the same distance, so
3x = 3.5(x-10)
3x = 3.5x - 35
35 = .5x
70 = x

The first car went 70mph and traveled 210 miles.
The second car went (70-10) = 60mph and traveled 60*3.5 = 210 miles.

Answer 10

Distance = Rate x Time (d = rt)

Distance for slow car: r(3.5)

Distance for fast car: (r + 10)(3)

Distances are the same

r(3.5) = (r + 10)(3)
3.5r = 3r + 30
.5r = 30
r = 60 {slow car}
r + 10 = 70 {fast car}

Use either rate & time to get distance

d = 60(3.5) = 210

or

d = 70(3) = 210