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54 Kmh-1 বেগের একটা গাড়ি 15 ms-2 ত্বরণে চলছে।50m দূরে এক বালককে দেখে গাড়িটি ব্রেক করে বালকটির 2m সামনে থেমে যায়.t=?

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asked Jan 28, 2014 in Physics by SNK (147 points)
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answered Feb 19, 2014 by saifur26 (126 points)

এখানে,

আদিবেগ, u=54 kmh-1

            =(54×1000 m)÷(60×60 s)

           = 15 ms-1

শেষবেগ, v = 0 ms-1

ত্বরণ, a = 15 ms-2

দূরত্ব, s = (50-2) m

         = 48 m

সময়, t = ?

আমরা জানি,

s = {(u+v)÷2}×t

বা, t = 2s÷(u+v)

বা, t  = 2×48 m ÷ (15 ms-1+ 0 ms-1)

 বা, t = 6.4 s(Ans.)

অতএব, সময়, t = 6.4 s (Ans.)

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answered Feb 5, 2014 by Fuad Hasan Wasi (120 points)
S={(v+u) ÷2}t

=>t=2s÷(u+v)

.     =2*48m÷(0+15) ms'1

.      =6.4s}

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