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A particle moves along a straight line such that its velocity V m/s from a given point is V=3t2 − 10t +3 where t is time in seconds. Find; 

  1. Acceleration of the particle when t=2 seconds
  2. The time taken to reach the maximum height 
  3. The distance covered during the 5th second
  4. Maximum velocity attained

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  1. Acceleration
    v=3t2 − 10t + 3
    a=dv/dt=6t−10
  2. time taken
    3t2 − 10t + 3=0
    3t2 − 9t −t+ 3=0
    3t(t−3)−1(t−3)=0
    (t−3)(3t−1)=0
     t=3s
     t=1/3s
  3. distance covered during 5th second

    =(125−125+15)−(64−80+12)
    =15−(76−80
    =15−−4
    =19metres
  4. maximum velocity
    Acceleration=0
    6t−10=0
    6t=10
    t=10/
     t=5/3 =12/3sec
    v=3(5/3) − 10(5/3) +3
    v=8.333−16.67+3
      = −5.34m/s
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1.6t-10
2.3 or 1/3seconds
3.19metres
4.-5.34m/s
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