Question

A train , travelling at 20km/h is approaching a platform . A bird is sitting on a pole on the platform .When the train is at a distance of 2km from pole,brakes are applied which produce a uniform declaration in it .At that instant the bird flies towards the train at 60 km/h and after touching the nearest point on the flies back to the pole and then flies towards the train and continues repeating itself.Calculate how much distance the bird covers before the train stops?​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial Speed of the train (u) = 20Km/h.
• Final speed of the train (v) = 0 km/h.
• Speed of the bird (v') = 60 km/h.
• The distance between Train and poll (s) = 2km.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Applying the Third Kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

Substituting the values,

$\large{\tt \leadsto (0)^2 - (20)^2 = 2 \times a \times 2}$

[Here we have taken final velocity as Zero. Because the question specifies that at the last train comes to rest.]

Simplifying ,

$\large{\tt \leadsto 0 - 400 = 4 \times a}$

$\large{\tt \leadsto - 400 = 4 \times a}$

$\large{\tt \leadsto a = \dfrac{-400}{4}}$

$\large{\tt \leadsto a = \dfrac{\cancel{-400}}{\cancel{4}}}$

$\large{\boxed{\tt a = - 100 \: Km/h^2}}$

So, the acceleration (deceleration) of the train is - 100 Km/h².

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Applying First kinematics equation,

$\large{\boxed{\tt v = u + at}}$

Substituting the values,

$\large{\tt \leadsto 0 = 20 + (- 100) \times t}$

$\large{\tt \leadsto 0 = 20 - 100t}$

$\large{\tt \leadsto 100t = 20}$

$\large{\tt \leadsto t = \dfrac{20}{100}}$

$\large{\tt \leadsto t = \dfrac{\cancel{20}}{\cancel{100}}}$

$\large{\boxed{\tt t = \dfrac{1}{5} \: hours}}$

So, the time taken by the train to come to rest is 1\5 hours.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Let the bird covers a Distance "D".

Now,

$\large{\boxed{\tt Distance = Speed \times time}}$

$\large{\tt \leadsto D = v' \times t}$

Substituting the values,

$\large{\tt \leadsto D = 60 \times \dfrac{1}{5}}$

$\large{\tt \leadsto D = \cancel{60} \times \dfrac{1}{\cancel{5}}}$

$\large{\tt \leadsto D = 12 \times 1}$

$\huge{\boxed{\boxed{\tt D = 12 \: Km}}}$

So, the bird travels a distance of 12Km.

$\rule{300}{1.5}$