Variable accerleration?

The acceleration a ms-2 of a particle after t seconds is given by a=4t^3−8

If the particle starts from rest, it is next stationary after _s

2 Answers
Shwetank Mauria
Apr 20, 2018

Particle is next stationary after #2# sec.

Explanation:

As acceleration is given by #a=4t^3-8#,

its velocity is given by #v(t)=int(4t^3-8)dt#

= #t^4-8t+c#

As the particle starts from rest, at #t=0#, its velocity is #0#

Hence #v(0)=0^4-8*0+c=0# i.e. #c=0#

Hence #v(t)=t^4-8t#

This will be #0# when #t^4-8t=0#

i.e. #t(t^3-8)=0# or #t(t-2)(t^2+2t+4)#

Observe that there are no real roots for #t^2+2t+4=0# and hence

#v(t)=0# when either #t=0# or #t=2#

Hence, particle is next stationary after #2# sec.

Narad T.
Apr 20, 2018

THe particle is stationary after #t=2s#

Explanation:

The acceleration is

#a(t)=4t^3-8#

The speed is the integral of the acceleration

#v(t)=int(4t^3-8)dt=t^4-8t+C#

Plugging in the initial conditions

#v(0)=0#

#0=0-0+C#

#C=0#

Therefore,

#v(t)=t^4-8t#

The particle is stationary when #v(t)=0#

That is,

#t^4-8t=0#

#t(t^3-8)=t(t-2)(t^2+2t+4)=0#

Therefore,

#{(t=0),(t-2=0),(t^2+2t+4=0):}#

#<=>#, #{(t=0),(t=2),(t=(-2+-sqrt(4-4*1*4))/(2)=(-2+-i2sqrt(3))/(2)=-1+-sqrt3):}#

We keep only the solution #t=2#

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