1.

The acceleration a in ms^-2 of a particle is given by a=3t^2+2t+2, where t is the time. If the particle starts out with a velocity v=2ms^-1 at t=0, then find the velocity at the end of 2s.

Answer»

Solution :`a=(dv)/(DT)=(3t^(2)+2t+2)dt`
`dv=(3t^(2)+2t+2)dt`
`int dv=int (3t^(2)+2t+2)dt`
`v=t^(3)+t^(2)+2t+C`
`c =2m//s, v=18 m//s" at " t=2s`.


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