Advanced Alegbra help?

Question

i never realy understood how to get these type of questions can anyone help me

Two cyclists start biking from a trail’s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

thanks

Answer 1

Qwyrx has a good way of thinking about it.

Here's another way.

The basic equation is d = rt.

In this case, since we're trying to find when they pass, the distance will be the same for both. So what we're really doing is setting the distance biker 1 travels, r1t1, equal to the distance biker 2 travels, r2t2.

If we say that t is the amount of time after the first bikers leaves (t1 = t), then the second biker is going to be traveling three hours less than t (t2 = t - 3).

And we're given the rates: r1 = 6, r2 = 10.

So...

r1t1 = r2t2
6(t) = 10(t - 3)
6t = 10t - 30
-4t = -30
t = 30/4 = 15/2 = 7.5 hours

But the problem asks how long after the second biker starts, so that's:

t - 3 = 7.5 - 3 = 4.5 hours.

Answer 2

The general pertinent formula is
Distance = Rate x Time (where you must keep the right side factors compatible. Note that this general formula can be written 2 other ways to apply to similar problems.)

To me it seems most plausible to equate the distances traveled by each biker:

2nd cyclist travels: 10 T, where T is the time in hours, the clock starting when he starts;

The first cyclist has a head strart 6(3) and using the one
clock we're watching, he continues 6T further.

The equation you have then is 6(3) + 6T = 10T
Solving for T you get 4.5 (the # of hours it takes for the
2nd cyclist to catch the cyclist who left first. Each traveled 45 miles.)

Answer 3

What you ned to do is translate the words into an algebraic equation.
The basic formula for time (t), distance (d) and speed (s) is: d = st
Then use t1, d1, s1 for the first cyclist
and t2, d2, s2 for the second cyclist
The next step is to write down as much as you know from what the wording says:
- When the 2nd cyclist catches with the first cyclist, they will both have travelled the same distance. Therefore, d1 = d2
- s1 = 6 mi/hr while s2 = 10 mi/hr
- the first cyclist starts 3 hours ahead. Therefore, when the 2nd cyclist catches up, hit time will be t1 = t2 + 3 hrs

The 3rd step is to create the equation. We know that d1 = d2 which means that s1 * t1 = s2 * t2. Let's substitute as much as what we know in the latter equation with the goal of have as few unknown variables as possible:
6 mi/hr * (t2 + 3 hrs) = 10 mi / hr * t2
As you can see, there is only one unknown variable, t2. There fore, you can solve for t2 using algebra. Once t2 is solved, you can go back to the data you gathered above and solve for the remaining variables.

Answer 4

Figure out how far apart they are when the second cyclist starts. Figure out how long it takes the second cyclist to travel that distance at a relative speed of 4 mph (the difference in their speeds, i.e., the speed at which the second cyclist is "catching up").

Answer 5

remember v = S / t or S = v.t
for cyclist number one, call it V1, S1,and t1.
cyclist number two , call it V2, S2,and t2.

because cyclist 2 started later, then t2 = t1 - 3 hours.

when cyclist 2 catched up cyclist one, it is when S1 = S2
v1.t1 = v2.t2
6 miles/hour . t1 = 10 miles/hour . (t1 - 3hours)
6 . t1 = 10 t1 - 30
- 4 t1 = -30
t1 = -30/(-4)
t1 = 7.5 hours.

remember t2 = t1 - 3
t2 = 7.5 - 3
t2 = 4.5

4.5 hours will pass before the second cyclist catches up with the first from the time the second cyclist started biking.

Answer 6

distance travelled d= vt , where v is the velocity and t is the time passed.
For 1st cyclist , d1 = 6 (t+3)
For 2nd cyclist, d2 = 10t .
When they meet,the distance should be the same ==>d1=d2
==>6(t+3)=10t ==> 18 = 4t ==> t=18/4 = 4.5hrs.

Answer 7

well you know speed=distance/time
they will all cover the same distance
therefore distance = speed*time

let the time spend by the first cylist be x
the the second cylist will be = x-3
the distance covered by the 1st cylist =6*x
for 2nd cylist=10*(x-3) and since they cover the same distance

6x=10(x-3)
6x=10x-30
10x-6x=30
4x=30
x=30/4
x=7.5
therefore at 7.5 miles the first and second cylist will be at the same spot.

Answer 8

x = time of first cyclist
x - 3 = time of second cyclist

10(x - 3) = 6x
10x - 30 = 6x
4x = 30
x = 7.5

so 4.5 hours

Answer 9

http://www.math.com the site will basically tutor you for free, very easy to understand and easy to navigate.