Question

Find time when boy catches the girl or when they are at their closest separation..

Answer 1

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

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

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

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

According to given question :

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