Question

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Answer 1

Case-1:

Velocity = 15 m/s

Time = 5 seconds

∴ Distance covered = (15*5)

                                = 75 m

Case-2:

Velocity = 5 m/s

Time = 10 seconds

Distance covered = (5*10)

                              = 50 m

Since, in the second time the boy turns back and start to walk again so, to find the displacement we have to subtract the distance covered in the second case from the distance covered in the second case.

∴ Displacement = (75-50)

                          = 25 m

Hence, the displacement is 25 m.

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Answer 2

Answer:

hey mate your answer is..

Explanation:

$condition \: 1) \: v 1= 15ms \\ t1 = 5sec \\ s1 = \frac{v1}{t1} \\ s1 = \frac{15}{5} \\s1 = 3m......(1). \\ condition \: 2 \\ v = 5ms \: and \: \: t = 10sec. \\ s2 = \frac{v2}{t2} \\ s2 = \frac{5}{10} \\ s2 = 0.5m.......(2) \\ from \: 1and \: 2 \\ s1 + s2 \\ = 3 + 0.5 \\ 3.5m..... \\ displacement \: is \: 3.5m.... \\ hope \: its \: helps........$