How much start A should give B so that the race ends in dead heat?

Question

A and B have a race on a circular track of 510 meters.A who runs faster meets with B in the middle of the 5th round.
Now in a 3 kM race how much start A must give B so that the race ends dead heat, i.e. A and B reach the goal at the same time.

Answer 1

what do you mean by the fifth round, do you mean the fifth lap?

I am interpreting your question to mean that A completes 4.5 laps when B completes 3.5 laps. This means that A travels 45/35 or 9/7 faster than B.

Thus, we can write the distance equations for A and B as:

A must travel x+3 km in the same time as B travels x km; A travels at 9/7v and B travels at v, so:

x+3=9/7 vt
3=vt =>t=3/v

substitute this expression for t into the first equation and get:

x+3=9/7v*(3/v)=27/7
x=27/7-3=6/7 km

A must start 6/7 km farther back; or A must travel 9/7 farther than B to finish at the same time, and 9/7x 3 = 3 6/7 km

Answer 2

Assuming the 5th round means the 5th lap...
A runs 510 meters * 4.5 laps.
At this point, A meets B.
Since this is the first time for A to pass him, B must have run exactly 1 lap less than A.
Thus, B has run 510 meters * 3.5 laps.

So, A runs 2295 meters in the same time B runs 1785 meters.

B runs 1785/2295 = 7/9 the speed of A.

So, put them at the finish line, grab a camcorder, and tell them both to run BACKWARDS. Stop recording when A, who is faster, finishes.
A ran 3000 meters, so in the same time B run 7/9 * 3000 = 2333.333333.... meters.

They looked funny running backwards, and if you play the tape in reverse, they look funny running forwards, but at least they're going the right way, AND doing exactly what you want them to do: B has a head start, and they both reach the finish line at the same time.

Since they are 3000 - 2333.33333333333 meters apart after A has run 3km, B needs a 666.666 meter head start (the devil's head start, so to speak) in order for them to finish a 3km race at the same time.

On second thought, 666.666 is disturbing. A 2000/3 meter head start sounds better. Or "six hundred sixty six and two thirds of a meter".

Answer 3

A very interesting question !

Let us assume that a's speed = x metres / second.
Let us assume that b's speed = y metres / second.

Let us assume that A gives B a headstart of w metres.

So , over a 3 km. race , if they finish together , we have the following equation :

3000/x = (3000 - w)/y

You have also mentioned that over a race on a circular track , A meets B in the middle of the 5th round. It is not clear whether A is in his fifth lap or B is in his fifth lap.

I assume B has completed 4 laps , and is in his fifth lap. So naturally A must have already completed 5 laps , and must be in his 6th lap.

The distances covered by the two of them are therefore :

A -> 510 * 5 + 510 / 2 = 2805 metres

B -> 510 * 4 + 510/2 = 2295 metres

Since they have met , they have completed these different distances in the same time.

So we can equate 2805/x = 2295/y

Solving these two equations , we get the headstart as 545.455 metres.

Answer 4

VA : speed of A
VB : speed of B
The speeds are stable aren’t they?
tA : time of A (to do circular distance)
tB : time of B (to do circular distance)
X is distance

X (which A meets B) = 510 × 4 + 510 / 2 = 2295 m

Let us assume VA = 2295m/s → tA = 1 s
VB = 1 m/s → tB = 2295 s
(you understand from there, A started race after B went 2294m )

Now distance is 3000 m
How much time A finish 3000m?
tA = (distance/speed) = 3000m/ 2295 m/s = (3000/2295) s (this is closely 1,307 s )

And how much time B finish 3000m?
tB = (distance/speed) = 3000m/1m/s = 3000 s

How much time B started before from A?
TB(we can call) = tB-tA = [3000-3000/2295 ] sn


This mean;
B make XB distance before A start,
XB = VB×TB = 1m/s × [3000-3000/2295 s] = 2998,6928104 m

Thus, B must start closely 2998,7 m before than A
so A and B reach the goal at the same time.

Answer 5

Distance of track = 510m.

For A:
Distance = 4∙5 * 510
Distance = 2295m

For B:
→
Distance = 3∙5 * 510j
Distance = 1785m

So the ratio of the speeds are:
B/A : 1785/ 2295
1 : 1∙2857.......

1∙2857... Ratio speed = 3Km (÷ 1∙2857...)
1∙2857.. /1∙2857 Ratio speed = 3Km/ 1∙2857...
1 Ratio speed = 2∙3333.... km
This is the distance that B has to run, so subtract it from the total distance to get the head start.
Head Start Distance of B = 3Km - 2∙3333.... Km.
Head Start Distance of B = 0∙3333...Km.

Answer 6

510m/(9/2)round=1020m/9round
That means that for every 510 m A travels, B behind A 1020/9 m.
3000m=3 km
510m slow by 1020/9 m
3000m slow by ? m
3000/510*1020/9=2000/3 m
So the answer is A behind B 2000/3 m or 666 and 2/3 m to start.

Answer 7

I wouldnt know Im only in the fourth grade!

Answer 8

Somehow, i believe the forumula :

Distance = Rate * time is involved or D = RT

or it's variants:

T = D/R
R=D/T

Good luck!

Answer 9

Okay, A meets B at 5.5 round, but u havent mentioned what lap B was on? 4.5/3.5/2.5/1.5/0.5 ?