my first post

It is my first post i am trying to post some of the questions and solutions for science and maths

Today's Question

If the position time relations is given by the expression x(t)=2t3+ 5t2 then find the velocity and accelaration at t=2 sec also comment is the velocity of the particle zero at any instant

velocity v =dx/dt v=d(2t3+ 5t2)/dt

v=6t2+10t

Velocity at t=2 sec

v=6x4+10x2

velocity=24+20=44 units

accelaration=d2x/dt2=dv/dt= 12 x t+10at t=2 sec

accelaration=24+10=34 units

If velocity=0 then6t2+10t=0

ie t(6t+10)=0

ie at t=0 or 6t+10=0

6t=-10

t=-10/6

Since time can not be negative . . . velocity is zero at t=0

ie at the start only

 

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