Question

A train is approaching a platform with a speed of $20\text{km/hr}$. A bird is sitting on a pole at the platform. When train is $2\text{km}$ away from the pole, brakes are applied so that the train decelerates uniformly, simultaneously the bird also flies towards the train with a velocity $60\text{km/hr}$. It touches the nearest point on the train and flies back and back again and so on. The total distance travelled by the bird before train stops is

• A. $30\text{km}$
• B. $15\text{km}$
• C. $12\text{km}$
• D. $10\text{km}$

Answer 1

The correct option is C $12\text{km}$

Let the magnitude of deceleration be d
Applying equation of motion on the train,
$v=u+at$
$0=20-dt$ -------------(1)
and $s=ut+1/2a{t}^2$
$2=20t-1/2\times d\times t^2$ -------------(2)
Putting $dt=20$ from equation 1 in equation 2,
$2=20t-10t$
$t=\frac{1}{5}\text{hr}$

Hence the bird flies with a speed of $60\text{km/hr}$ for a time of $\frac{1}{5}\text{hr}$
Hence distance traveled by the bird = $60\times \frac{1}{5}=12\text{km}$