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Two (chicken race) runners travel against each other with the (constant) velocity of 10 km/h each. The distance between them is 50 m. In how many seconds will they crash into each other?

Full calculation needed to get full marks.

HolyKonni wrote:

Two (chicken race) runners travel against each other with the (constant) velocity of 10 km/h each. The distance between them is 50 m. In how many seconds will they crash into each other?

Full calculation needed to get full marks.

They both have to travel a distance of 25 m (50 m/2=25 m).

10 km=10 000 m
1 h=60 min=3600 s
10 000 s/3600~2,78 s

Each sprinter travels with a constant velocity of 2,78 m/s. The time it takes to travel 25 m with a velocity of 2,78 m/s is: 25 m/(2,78 m/s)~8.99 s

So my answer is 8.99 seconds.

Did I win? I hope this isn't one of those faggot ass questions where the answer is like "They never crash because one of them stops to eat pizza and the other one fucks his grandmother, blablabla I'm a retard blablabla..." Because I swear to god, I will kill you like 3 hours!

Why pull my finger, you certainly are the third best kid in math in Sweden. Let's have a look at this.

BattleSmurf wrote:

They both have to travel a distance of 25 m (50 m/2=25 m).

true, they have to travel a distance of 25 m each.

BattleSmurf wrote:

10 km=10 000 m
1 h=60 min=3600 s
10 000 s/3600~2,78 s

ok, ok. here you transform km/h into m/s. you can also just do it like this:
10km/h / 3,6 = ~ 2,78m/s. remember:
- from km/h to m/s you divide by 3,6
- from m/s to km/h you multiply with 3,6

BattleSmurf wrote:

Each sprinter travels with a constant velocity of 2,78 m/s. The time it takes to travel 25 m with a velocity of 2,78 m/s is: 25 m/(2,78 m/s)~8.99 s

true. becouse v=s/t

BattleSmurf wrote:

So my answer is 8.99 seconds.

that is very correct.

BattleSmurf wrote:

Did I win?

congrats, yes you did win. but you didn't get full marks, though sorry. how you could've gotten full marks:
- you should have clarified that you transformed from km/h to m/s.
- you could have shown that you used the formula v=s/t
- you should have answered in the correct significant figures, becouse there are 1 or 2 sig. fig in 10 km/h and 50 m, you should have answered in 1 or 2 sig. fig. form. so that the answer is 8.9 seconds, 9.0 seconds (2 sig. fig.) or just 9 seconds(1 sig. fig.)


otherwise great calculation.

This is how I did it:

50 m = 0,05 km

Each of the runners has a vel. 10km/h. Both of them added together have 20km/h. When I do it this way I don't divide the distance, I add the velocities instead.

v=delta s/ t (constant velocity = change in displacement / change in time)

20 km/h = 0,05 km / t

t = 0,05 km / 20 km/h = 0,0025 h

to transform hours into seconds you multiply by 3600 (60*60), so:

0,0025 * 3600 = 9 s.

Answer (in 1 sig. fig.): It took 9 seconds.

btw, sorry I have no price. you win honor instead.

HolyKonni wrote:

BattleSmurf wrote:

10 km=10 000 m
1 h=60 min=3600 s
10 000 s/3600~2,78 s

ok, ok. here you transform km/h into m/s. you can also just do it like this:
10km/h / 3,6 = ~ 2,78m/s. remember:
- from km/h to m/s you divide by 3,6
- from m/s to km/h you multiply with 3,6

But then you'll have to clarify why you divide it by 3,6

HolyKonni wrote:

congrats, yes you did win. but you didn't get full marks, though sorry. how you could've gotten full marks:
- you should have clarified that you transformed from km/h to m/s.
- you could have shown that you used the formula v=s/t
- you should have answered in the correct significant figures, becouse there are 1 or 2 sig. fig in 10 km/h and 50 m, you should have answered in 1 or 2 sig. fig. form. so that the answer is 8.9 seconds, 9.0 seconds (2 sig. fig.) or just 9 seconds(1 sig. fig.)


otherwise great calculation.

The formula v=s/t doesn't need to be part of the solution. "25 m/(2,78 m/s)~8.99 s" can easy be calculated in a regular way, because "m" and "s" doesn't have any values.

25m/(2,78m/s) = 25m/m/(2,78m/s)/m = 25/2,78s ~ 8.9s

And the part about the significant figures is correct, but if this problem was to be used on a test, it's usually explained how many decimals the answer should be in.