Question

Question no - 3



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Answer 1

AnSwEr:-

$\small\rightarrow \: A \: to \: B \: = 2 \: min \: in \: 30 \: sec$

$\rightarrow120 + 30 = 150 \: sec$

$\rightarrow \: B \: to \: C = 1 \: min = 60 \: sec$

$Total \: distance \: covered \: \: \\ from \:AB = 300 \: m$

$\rightarrow \: Total \: time \: taken \: = 2 \times 60 + 30 \: sec \: = 150 \: sec$

$\rightarrow \ \: Average \: velocity \: = \frac{total \: displacement}{ \: total \: time \: taken}$

$\rightarrow \: \frac{300}{150} = 2 \: m /s$

$\rightarrow \: Total \: distance \: covered \: from \: Ac = ab \: + bc \: = 300 + 200 \: m$

$\rightarrow \: Total \: time \: taken \: from \: a \: to \\ \: c \: = Time \: taken \: for \: \: Ab \: to \\ Time \: taken \: for \: bc \: \\ = (2 \times 60 + 30)60 \: sec \\ = 210 \: sec \:$

$\rightarrow \: Therfore. \: average \: speed \: from \\ \: ac = Total \: distance / \: total \: time \: taken \: \\ = 400/210 \: m {}^{ - 1}$

$\rightarrow \: Displacement \: (s) \: from \: A \: \\ \: to \: c \: = AB - Ac \\ = 300 - 100 \\ = 200 \: m$

$\rightarrow \: Time \: taken \: for \: displacement \ \\ from \: \: Ac \: = 210 \: sec \: . \\ Therefore \: velocity \: from \: \\ A \: to \: c \: = \: displacement$

$\rightarrow \frac{200}{210} = 0.952ms\ ^{ - 1}$