a = -4
v = -4t+c; v(2)=6, so c=14
v = -4t+14
s = -2t^2+14t+c; s(2) = 3, so c = -17
s = -2t^2 + 14t - 17
Now you can answer the questions
A particle residing in one dimension is moving with constant acceleration a=-4m/sec^2 .At time t =2sec, the particle is located at x=3m and is observed to have velocity v=6m/sec. (a) At what instants is the particle to be found at x=0m?(ii)what is the velocity of the particle at each of these instant? (b)(i)At what instant is the particle stopped ,i.e., v=0m/sec?(ii) where is the particle
2 answers
A particle residing in one dimension is moving with constant acceleration a=-4m/sec^2 .At time t =2sec, the particle is located at x=3m and is observed to have velocity v=6m/sec. (a) At what instants is the particle to be found at x=0m?(ii)what is the velocity of the particle at each of these instant? (b)(i)At what instant is the particle stopped ,i.e., v=0m/sec?(ii) where is the particle
1 answer