A body is moving with an initial velocity (v) from an initial position (#x_0#) the expression for the acceleration #a (t) =3t^2-4t^4#. Find the expression for x (t) and v (t)?

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Answer 1 · 2018-02-24T06:18:55

The expressions are #x(t)=1/4t^4-2/15t^6+vt+x_0# and #v(t)=t^3-4/5t^5+v#

#a(t)=3t^2-4t^4#

The velocity is the integral of the acceleration

#v(t)=inta(t)dt=int(3t^2-4t^4)dt#

#=3/3t^3-4/5t^5+C#

When #t=0#, #v(0)=v#

Therefore,

#v=0-0+C#

#v(t)=t^3-4/5t^5+v#

The position is the integral of the velocity

#x(t)=intv(t)dt=int(t^3-4/5t^5+v)dt#

#=1/4t^4-4/30t^6+vt+C_1#

The initial position is

#x(0)=x_0#

Therefore,

#x(0)=0+C_1=x_0#

So,

#x(t)=1/4t^4-2/15t^6+vt+x_0#