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

550 m

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137.5 m

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412.5 m

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275 m

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Solution

The correct option is C 412.5 m
u=0,v=27.5m/s and t=10 sec.
$∴ a = \frac{27.5 - 0}{10} = 2.75 m / s^{2}$
Now,the distance travelled in next 10 sec,
$S = u t + \frac{1}{2} a t^{2} = 27.5 \times 10 + \frac{1}{2} \times 2.75 \times 100$
=275+137.5=412.5 m

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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

The correct option is C 412.5 mu=0,v=27.5m/s and t=10 sec. ∴a=27.5−010=2.75m/s2 Now,the distance travelled in next 10 sec, S=ut+12at2=27.5×10+12×2.75×100 =275+137.5=412.5 m

The correct option is C 412.5 m
u=0,v=27.5m/s and t=10 sec.
$∴ a = \frac{27.5 - 0}{10} = 2.75 m / s^{2}$
Now,the distance travelled in next 10 sec,
$S = u t + \frac{1}{2} a t^{2} = 27.5 \times 10 + \frac{1}{2} \times 2.75 \times 100$
=275+137.5=412.5 m