Two cars race around a circular track, in opposite directions, at constant rates. They start at the same position and meet every 30 seconds. If they move in the same direction, they meet every 120 seconds. If the track in 1800 meters long, what is the speed of each car?

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John M. | John - Algebra TutorJohn - Algebra Tutor

Givens:

2 cars

Travel distance 1800 meters

Variables and Definitions:

Velocity(V) is speed in meters/second (m/s)

Time(t) is in seconds (s)

a)Find the speed of the cars if they travel in opposite directions on the track and pass each other every 30 seconds.

Since the cars are traveling opposite each other, we add their speeds. In 30 seconds the two cars together travel 1800 meters

30s(V1m/s+V2m/s)=1800m (note: I divide both sides by 30 because we want the speeds, not distances)

So we have one equation with 2 unknowns

(Eq. 1)V1+V2 = 60

b) Find the speed of the cars if they travel in the same direction on the track and pass each other every 120 seconds.

Since the cars are travel the same direction, we subtract their speeds. In 120 seconds the difference in distance travelled = 1800 meters.

120s(V1m/s-V2m/s) = 1800

We now have a second equation with 2 unknowns:

(Eq. 2) V1-V2=15

To find V1 and V2 we need to solve the system of the two equations, 1 and 2.

V1+V2=60

V1-V2=15

If we add these two equations we get: 2V1=75, so V1=37.5. Straighforward substitution is all it takes to find V2.

V1=37.5m, V2=22.5m

Let the two cars have a speed u and a speed v, respectively.

(u+v)30 = 1800 =>u+v = 60 ......(1)

(u-v)120 = 1800 => u-v = 15 ......(2)

(1)+(2): 2u = 75=> u = 37.5 m/s

=> v = 60-u = 22.5 m/s

Hey Olivia -- head on, the cars combined speeds cover 1800m/30s => 60m/s combo

... in same direction, the speed diff (dV) takes 4x as long (120s/30s) => dV = 15m/s

... start at 30m/s midpoint and +/- 7.5m/s ==> *fast car 37.5 m/s, slow car 22.5 m/s* :)

Hi Olivia, Let x = the speed of Car 1 and let y = the speed of Car 2. Then...

Eq 1...30x + 30y = 1800m (This is how far the cars will travel in 30 seconds together).

Eq 2...120x - 120y = 1800m (This is how much farther car 1 will travel to lap car 2 in 120 seconds).

Now multiply equation 1...by 4 to get equation 3 below. Then add equations 3 and 2 to get equation 5.

Eq 3...120x + 120y = 7200m

Eq 2...120x - 120y = 1800m

Eq 5...240x = 9000m x = 9000/240 = 37.5 m/s (Car 1 speed)

Now subtract equation 2 from equation 3 to get equation 6.

Eq 6...240y = 5400m y = 5400/240 = 22.5 m/s (Car 2 speed)

This is an interesting problem, just needs a little imagination to solve it. The basic logic is this: Whey they are going in opposite directions, they meet again when they both TOGETHER cover the whole lap. When they are going in the same direction they meet the second time (and every time thereafter) when the faster car completed a full lap and then covered the extra distance that the slower car covered in its first lap yet.

Suppose the speeds of the cars A and B are a kmph and b kmph (kilometers per hour) respectively, b being faster than a.

When they are driving in opposite directions, they are moving away from each other and in 30 seconds car A covers a * 30/3600 km. In that same time car B covers b * 30/3600 km. Since they meet at the end of 30 seconds, they together covered the full lap of 1800 meters (1.8km)

(a + b) / 120 = 1.8

When they going in the same direction, car A covered a * 120/3600 km and car B covered b * 120/3600 km. Since we said car B is faster, and it caught up with car A in its second lap, we can form the following equation

(b / 30)=1.8 + (a / 30)

or (b - a) / 30 = 1.8

We form the two equations

a + b = 120 * 1.8 = 216

b - a = 30 * 1.8 = 54

2b = 270 or b = 135 kmph

a = 216 - b = 216 -135 = 81kmph

Let's first find the angular speeds of each car, ω_{1} and ω_{2, }
in degrees per second. The two cars race through angles θ_{1}=ω_{1}t and θ_{2}=ω_{2}t.

If they race in opposite directions they meet when t=30 s and θ_{2}=360°-θ_{1}, which tells us

ω_{2} (30) = 360 - ω_{1} (30) , or

ω_{2} = 12 - ω_{1}.

If they race in the same direction they meet when t=120 s and θ_{2}=360°+θ_{1}, which tells us

ω_{2} (120) = 360 + ω_{1} (120) , or

ω_{2} = 3 + ω_{1}.

You have two equations for ω_{1} and ω_{2}; solve them and get

ω_{1} = 4.5 °/s and ω_{2} = 7.5 °/s

Now use the fact that 360° corresponds to an 1800-m circumference, so 4.5° corresponds to an arc of 4.5/360*1800=22.5 m and 7.5° to 37.5 m.

Therefore, the two speeds are

Olivia, the key here is to create an image of what is happening, and then translate that image into two equations with two unknowns (the velocities of the cars).

So, Let's say car #1 has a velocity V1, and car #2 has a velocity of V2

When they travel in opposite directions and meet every 30 seconds, this means that together the cars cover the entire 1800 meter track in 30 seconds, or that their combined velocity (V1+V2) has to equal the length of the track divided by the time it took to cover it (1800/30=60 meters per second). That's our first equation (V1+V2=60).

When they travel in the same direction, the equation is a little trickier. To visualize what is happening, let's assume that car 1 is faster i.e. V1>V2 (you can choose either one), and imagine what happens. Car 1 jumps out to a lead and keeps pulling away from Car 2, until it comes up from behind and "laps" Car 2 by passing it, 120 seconds after starting out. So however far Car 2 has traveled, the Car 1 has to have traveled an additional 1800 meters (1 additional lap). The distance Car 2 has traveled is equal to its velocity (V2) times the time it traveled (120 seconds) [120xV2]; the distance Car 1 has traveled is also equal to its velocity (V1) times the time it traveled (120 seconds) [120xV1], and it also has to equal how far Car #2 traveled plus 1800 meters of the extra lap [120xV2+1800], so our second equation is

120xV1=120xV2+1800.

120xV1=120xV2+1800.

So, you now have two equations and two unknowns, which I leave you to solve

V1+V2=60

120xV1=120xV2+1800

Let me know if you need any help solving 2-equation-2-unknown problems, and I would be happy to walk you through that as well. John

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