An object accelerates from rest to a velocity 27.5 ,m/s in 10 ,sec then find distance covered by obj

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2.Motion in Straight Line

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An object accelerates from rest to a velocity $27.5 \,m/s$ in $10 \,sec$ then find distance covered by object in next $10\, sec$........$m$

A

$550$

B

$137.5$

C

$412.5 $

D

$275 $

Solution

(c) $u = 0$, $v = 27.5\;m/s$ and $t = 10 \,sec$

$\therefore a = \frac{{27.5 – 0}}{{10}} = 2.75\;m/{s^2}$

Now, the distance traveled in next $10\, sec$,

$S = ut + \frac{1}{2}a{t^2}$$ = 27.5 \times 10 + \frac{1}{2} \times 2.75 \times 100$

$= 275 + 137.5 = 412.5 \,m$

Standard 11

Physics

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