A bird, 720 km away from a train flies towards it at a speed of 120km/h. As soon as it reaches the train, it turns back towards the wall (starting position). The train travels at a speed of 15m/s towards the starting position of the bird. How much distance towards the wall will the bird cover before the train reaches it.

I tried solving this problem by relative speed concept like this: Ratios of velocities (towards:away)=29:11 Thus the ratio of time taken in travelling towards the train to away from the train =11:29 Thus, the bird spent a total of 29/( 29 + 11) × 13.33 hours travelling away from the train. The velocity of the bird travelling away from the train is 120km/h Thus, the distance travelled by the bird away from the train is 13.33 × 29/40 × 120 = 1160 km. But the Text book answer comes out to be 800 Km.Please someone explain me..

  • $\begingroup$ This has nothing to do with linear algebra. $\endgroup$ – Kenny Lau May 15 '17 at 9:02
  • $\begingroup$ Sorry I am Unaware of it $\endgroup$ – Roy May 15 '17 at 9:03

The time of interest is (half) the time it takes the train to reach the bird's starting position ($x_0$). The train's veolcity is $15 m/s = 54 km/h$, and will thus take $T=13 \frac{1}{3}$ to reach $x_0$.

After every occasion the bird meets the train (It does not matter where) it (bird) just does the same distance back to her $x_0$. Since we want the total traveled distance of the bird towards $x_0$, it follows we want the distance the bird may travel when given $T/2 = 6 \frac{2}{3}$ hours with velocity $V = 120 km/h$.

$D=vt=\frac{VT}{2}=120 \times6 \frac{2}{3}= \frac{120 \times 20}{3}=800 km/h$

  • $\begingroup$ I am unable to understand why and how have u assumed T/2 $\endgroup$ – Roy May 15 '17 at 9:50
  • $\begingroup$ The bird changes "direction" at $2$ types of occasions. The first is upon reaching $x_0$. The 2nd is upon reaching the train. Whenever she meets the train, she turns right back, and will do the same distance (in the opposite direction) back to $x_0$. We know her velocity is constant. Clear enough? $\endgroup$ – user76568 May 15 '17 at 9:54
  • $\begingroup$ OK got it.Thank u :-) $\endgroup$ – Roy May 15 '17 at 9:56
  • 1
    $\begingroup$ You are welcome. $\endgroup$ – user76568 May 15 '17 at 9:57

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