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kupik [55]
2 years ago
10

A train leaves the station at 11 a.m. traveling at the rate of 40 miles per hour. A faster train leaves the same station at 1 p.

m. that afternoon and travels in the same direction on a parallel track at a rate of 60 miles per hour. At what time will the faster train overtake the slower one
Mathematics
1 answer:
Reika [66]2 years ago
6 0

The faster train overtake the slower on 5 p.m.

How to solve such time related questions?

The key concept for solving such questions is the relationship between speed, distance and time.

The relationship between them is Distance = Speed x Time.

Let the number of hours after 1pm at which faster train will overtake slower train be x.

At overtaking point,

Distance travelled by slower train = Distance travelled by faster train

Given :

  • Speed of slower train = 40 miles per hour
  • Speed of faster train = 60 miles per hour
  • Time taken by faster train = x
  • Time taken by slower train = x + 2.

On equating distances we get

40 x ( x + 20) = 60 x (x)

40x + 80 = 60x

20x = 80

x = 4.

∴The faster train overtake the slower on 1 + 4 = 5 p.m.

Learn more about time related questions here  :

brainly.com/question/23708432

#SPJ4

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Step-by-step explanation:

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It is impossible these two tangents intersect each other outside the circle because they parallel to each other

Step-by-step explanation:

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- This point is called the point of tangency.

- The tangent to a circle is perpendicular to the radius at the point of

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- The tangent to a circle is perpendicular to the diameter at one of its

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* Now lets solve the problem

- If we have a circle with center O

- AB is a diameter of circle O

- CD is a tangent to circle O at point A

- EF is a tangent to circle O at point B

∵ AB is a diameter

∵ CD is a tangent to circle O at A

∵ The tangent and the diameter are perpendicular to each other at the

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