Question

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 5^th second
4. Maximum velocity attained

Answer 1

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=¹⁰/₆ 
    t=⁵/₃ =1²/₃sec
   v=3(⁵/₃) − 10(⁵/₃) +3
   v=8.333−16.67+3
     = −5.34m/s

Answer 2

1.6t-10
2.3 or 1/3seconds
3.19metres
4.-5.34m/s