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PLEASE ANSWER ASAP(TEST TODAY!)
Acceleration of a particle moving rectilinearly is a=4-2x.It is at instantaneous rest at x=0.At what position x(in m)will the particle again come to instantaneous rest?

MANAS GOEL , 6 Years ago
Grade 11
anser 1 Answers
HIMANSHU SHEKHAR DAS

Last Activity: 6 Years ago

as the relation is give in x so we have to find the differential relation between a, dv,dx
we know that acceleration is equal to rate of instantaneous change of velocity.
so a=dv/dt
multiplying dx in both numerator and denominator, we get a= (dv/dt)(dx/dx)
we know that dx/dt is nothing but velocity so we get a important relation that a=v*dv/dx
so v*dv/dx=4-2x
=> vdv=4-2x
integrating both sides and putting appropriate limits at x=0 the body was at instantaneous rest, so v=0 and let at a general velocity v  the position vector is x.
\int_{0}^{v}vdv = \int_{0}^{x}xdx
=> v2/2 = 4x – x2
=> v2= 8x – 2x2
we know that at instantneous rest the velocity is 0
 so v2  is also 0
=> 8x=2x2 => x=0 m or x=4 m
so the body comes to instantaneous rest again at x=4 m
thanks
 

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