Question

A particle starts from a point with a velocity of 6m/s and moves with an acceleration of -2m/s^2.show that after 6 s the particle will be at starting point

Answer 1

use s= ut+1/2at^2
Given s=0, u=6, a= -2
0 = 6t-1/2*2*t^2 or t(t-6) = 0
t= 0 or t=6 sec

Answer 2

Aɴꜱᴡᴇʀ

☞ Its true.

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Gɪᴠᴇɴ

$\tt \leadsto Initial \: velocity(u) = 6 \: m/s \\ \\ \tt \leadsto Acceleration(a) = - 2 \: {m/s}^{2} \\ \\ \tt \leadsto Time(t) = 6 \: sec$

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ᴛᴏ ꜰɪɴᴅ

❍ That the particle will be at the starting point after 6 sec.

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Sᴛᴇᴘꜱ

$\begin{lgathered}\bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2}a {t}^{2} \\ \\ \tt: \implies s = 6 \times 6 + \frac{1}{2} \times ( - 2) \times {6}^{2} \\ \\ \tt: \implies s = 36 - {6}^{2} \\ \\ \tt: \implies s =36 - 36 \\ \\ \tt: \implies s =0 \: m \\ \\ \green{\tt \therefore Displacement \: is \: 0 \: m} \\ \: \green{\tt So, \: after \: 6 \: sec \: particle \: will \: be \: on \: starting \: point} \\ \\ \blue{\bold{Some \: related \: formula}} \\ \orange{\tt \circ \:v = u + at} \\ \\ \orange{\tt \circ \: {v}^{2} = {u}^{2} + 2as}\end{lgathered}$

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