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1 A particle is moving with constant initial velocity4 ms-1 till t=1.5s. Then it accelerates at10 ms-2 from 1.5 s to 3 s. The distance coveredis(1) 17.25 m(2) 36.25 m(3) 40 m(4) 23.25 m |
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Answer» Explanation:Solution,Here, we haveInitial velocity of particle, U = 4 m/sAcceleration by particle, a = 2 m/sTime taken by particle, t = 1.5 sTo FIND,Distance covered, s = ?According to 2nd equation of motion,We KNOW that,s = UT + 1/2 at²So, putting all the values, we gets = ut + 1/2 at²⇒ s = 4 × 1.5 + 1/2 × 2 × 1.5²⇒ s = 6 + 1/2 × 2 × 2.5⇒ s = 6 + 1/2 × 5⇒ s = 6 + 2.5⇒ s = 8.5 mHence, the distance covered by particle is 8.5 m. |
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