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