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2 years ago

A train travels 81 kilometers in 2 hours and then 90 kilometers in two hours. wat is the average speed

1 answer:

5 0

I'm assuming your question refers to the train travels 81 km in 2 hours, as in that's the total distance covered versus the speed for those 2 hours, whereas the 90 km in 2 hours the second time around is the total distance not speed I would assume.

Now... if it was going 81 km for the first 2 hours and 90 km for the second 2 hours then the average speed would be the mean of these numbers, with that being 85.5 km. Though, I doubt that's your question.

With that said, 81 km covered by 2 hours and 90 km covered by 2 more hours. To acquire the km an hour average, we'll have to divide the distance by how many hours it traveled:
81 / 2 = 40.5
90 / 2 = 45

Meaning the train was going 40.5 km an hour for the first 2 hours and 45 km an hour for the second 2 hours. Now, to find the mean:
40.5 + 45 = 85.5 / 2 = 42.75

In-case you were wondering, the mean is the sum of all the numbers in a set, divided by the total amount of numbers in that set, take for instance:
3, 5, 9, 2, 1, 5. -> There are 6 numbers in this set.
3 + 5 + 9 + 2 + 1 + 5 = 25 -> 25 is the sum of these numbers.
25 / 6 = 4.2 (estimated) -> 4.2 is roughly the mean or average between the original set.

Anyhow, with that aside the average speed between this I would believe would be 42.75 km an hour. I hope that helps, have a great rest of your day! ^ ^
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3 years ago

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According to the law of conservation of momentum:

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v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
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Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

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2 years ago

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