Question

Solve the question in the attached image.​

Answer 1

QUESTION:

Two trains, each having a speed of 30 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the (crazy) bird flies directly back to the first train, and so forth. What is the total distance the bird travels before the trains collide?

SOLUTION:

Given:

• (v₁) = 30 km/h
• (v₂) = 60 km/h
• (d) = 60 km

Converting km/h to m/s and km to m,

(v₁) = 30 km/h

$\implies v_{1}=30 \times\frac{5}{18} = 8.33 ms^{-1}$

(v₂) = 60 km/h

$\implies v_{2}= 60\times \frac{5}{18}= 16.67 ms^{-1}$

(d) = 60 km

$\implies 60 \times 1000 = 60000 m$

• The train will collide in the middle,

Δx = 30 km

So,

When train collides,

$\boxed{v=\frac{\triangle x}{\triangle t}}$

Also,

$\boxed{\triangle t=\frac{\triangle x}{v}}$

Now,

Putting the given values, we get

$\implies \triangle t = \frac{30000}{8.33}=3601$

Now,

$\boxed{v=\frac{\triangle x}{\triangle t}} \implies \triangle x=v\triangle t$

So,

Δx = (16.67)m/s * (3601) ≈ 60028.67 m

Δx = 60 km (Approx)

Answer 2

Answer:

Given :-

• (v₁) = 30 km/h
• (v₂) = 60 km/h
• (d) = 60 km

To Find :-

total distance the bird travels before the trains collide

Solution :-

At First let's convert unit

30 × 5/18 = 150/18 = 8.33 m/s

60 × 5/18 = 300/18 = 16.67 m/s

60 = 60 × 1000 = 60000 m

Now,

We know that

∆t = ∆x/v

∆t = 3000/8.33

∆t = 360.18 ≈ 360

Now,

∆x = v × ∆t

∆x = 16.67 × 3601

∆x = 60028.6 m

∆x = 60,000 m or 60 km