Question

find time when boy catches the girl or when they are at their closest separation..
velocity of boy is 50m/s and velocity of girl is 30m/s and acceleration of boy is 1m/s^2 and acceleration of girl is 2m/s^2 and distance between them is 150m​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Time=10\:sec\:and\:30\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Velocity \: of \: boy = 50 \: m/s \\ \\ \tt: \implies Velocity \: of \: girl = 30 \: m/s \\ \\ \tt: \implies Acceleration \: of \: boy = {1 \: m/s}^{2} \\ \\ \tt: \implies Acceleration \: of \:girl= {2\: m/s}^{2} \\ \\ \tt: \implies Sepration \: between \: them = 150 \: m \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies Time \: taken \: to \: catch \: the \: girl =?$

• According to given question :

$\green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2} \\ \\ \tt: \implies 300 = 40t - {t}^{2} \\ \\ \tt: \implies {t}^{2} - 40t + 300 = 0 \\ \\ \tt: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} \\ \\ \tt: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} \\ \\ \tt: \implies t = \frac{40 \pm 20}{2} \\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}$