question_answer

Particles A and B are free to move along parallel paths. B is moving at a constant velocity of 10 m/s and A starts (from rest) moving at the moment B passes it. If A accelerates at \[1\,\,m/{{s}^{2}}\]for 5 s and then travels at a constant velocity, the distance between A and B 10 s after B passes A is ___. (a) 5 m                                  (b) 62.5 m (c) 12.5 m                            (d) 17.5 m

Answer:

Total distance travelled by particle B for 10s \[\tan \theta =\frac{1/2}{6}=\frac{1}{12}\] Distance travelled by particle A for 1st 5s, \[\theta ={{\tan }^{-1}}\left( \frac{1}{12} \right)\] \[{{\tan }^{-1}}\left( \frac{1}{12} \right)\] (\[XY=?\]\[XY=XZ+ZY=2XZ\text{ }or\text{ }2ZY\]) \[\Delta OZX,\] \[\sin \,60{}^\circ =\frac{XZ}{r}\] \[=\frac{r\sqrt{3}}{2}\] Distance between\[\therefore \]