Question

ibigay Ang kahalagahan

1.Hindi, si Kanor ang kumuha ng pulang manok

2.Hindi si Kanor ang kumuha ng pulang manok

3.Hindi si Kanor, ako ang kumuha ng pulang manok​

Answer 1

Explanation:

$Answer:</p><p></p><p>\green{\tt{\therefore{Time=10\:sec\:and\:30\:sec}}}∴Time=10secand30sec</p><p></p><p>\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}Step−by−stepexplanation:</p><p></p><p>\begin{gathered}\begin{gathered}\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 =? \end{gathered}\end{gathered}Given::⟹Velocityofboy=50m/s:⟹Velocityofgirl=30m/s:⟹Accelerationofboy=1m/s2:⟹Accelerationofgirl=2m/s2:⟹Seprationbetweenthem=150mToFind::⟹Timetakentocatchthegirl=?</p><p></p><p>• According to given question :</p><p></p><p>\begin{gathered}\begin{gathered} \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}\end{gathered}\end{gathered}⋆Using relative motion method∘Netvelocity=50−30=20m/s∘Netacceleration=1−2=−1m/s2Asweknowthat:⟹s=ut+21at2:⟹150=20×t+21×−1×t2:⟹300=40t−t2:⟹t2−40t+300=0:⟹t=2×1−(−40)±(−40)2−4×1×300:⟹t=240±1600−1200:⟹t=240±20:⟹t=10secand30sec</p><p></p><p>━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━</p><p></p><p>\sf\red{@MsHeaven}@MsHeaven</p><p></p><p>$

$Answer:</p><p></p><p>\green{\tt{\therefore{Time=10\:sec\:and\:30\:sec}}}∴Time=10secand30sec</p><p></p><p>\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}Step−by−stepexplanation:</p><p></p><p>\begin{gathered}\begin{gathered}\green{\underline{\bold{Given :}}} \\ \tt: \implies Velocity \: of \: boy = 50 \: m/s \\ \\ \tt: \implies Velocity \: of \: girl = 30 \: m/s \\ \\ \tt: \implies Acceleration \: of \: boy = {1 \: m/s}^{2} \\ \\ \tt: \implies Acceleration \: of \:girl= {2\: m/s}^{2} \\ \\ \tt: \implies Sepration \: between \: them = 150 \: m \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies Time \: taken \: to \: catch \: the \: girl =? \end{gathered}\end{gathered}Given::⟹Velocityofboy=50m/s:⟹Velocityofgirl=30m/s:⟹Accelerationofboy=1m/s2:⟹Accelerationofgirl=2m/s2:⟹Seprationbetweenthem=150mToFind::⟹Timetakentocatchthegirl=?</p><p></p><p>• According to given question :</p><p></p><p>\begin{gathered}\begin{gathered} \green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2} \\ \\ \tt: \implies 300 = 40t - {t}^{2} \\ \\ \tt: \implies {t}^{2} - 40t + 300 = 0 \\ \\ \tt: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} \\ \\ \tt: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} \\ \\ \tt: \implies t = \frac{40 \pm 20}{2} \\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}\end{gathered}\end{gathered}⋆Using relative motion method∘Netvelocity=50−30=20m/s∘Netacceleration=1−2=−1m/s2Asweknowthat:⟹s=ut+21at2:⟹150=20×t+21×−1×t2:⟹300=40t−t2:⟹t2−40t+300=0:⟹t=2×1−(−40)±(−40)2−4×1×300:⟹t=240±1600$

hope it helps uh