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Alex777 [14]
3 years ago
9

Two trains start from towns 224 mi apart and travel towards each other on parallel tracks. They pass each other 1.6 hr later. If

one train travels 10 mph faster than the​ other, find the speed of each train.
Physics
1 answer:
Darina [25.2K]3 years ago
3 0

Answer: 65mph, 75mph

Explanation:

Let us assume x to be the speed of the slower train, in mph (miles per hour).

Then the speed of the other train is (x+10) mph, according to the question.

We then would have an equation like this

1.6x + 1.6(x+10) = 224.

This is because, the first addend in the left side is the distance covered by the slower train.

The second addend in the left side is the distance covered by the faster train.

The sum is 224 miles, because they together covered all the distance to the moment when they meet each other.

1.6x + 1.6x + 16 = 224

3.2x + 16 = 224

3.2x = 224 - 16

3 2x = 208

x = 208/3.2

x = 65

Thus the speed of the slower train is 65mph, and that of the other train is 65 + 10 = 75mph

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a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

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This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

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         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

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           t_total = 2 t

           t_total = 2 v₀ sin θ / g

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