The velocity of particle moving along a straight line increases according to the law#v=v_0+Kx# , where K is a positive constant,then options are below?

Question

the acceleration of the particle is #K(v_0+Kx)#

the particle takes a time #(1/k)*ln(v/v_0)# to attain speed of v

velocity varies linearly with displacement with slope of velocity-displacement curve equal to K.

all of the above
~correct option given is 4(help me)

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Answer 1 · 2018-07-02T20:37:44

Above three options are correct

Given the velocity of particle as a function of displacement #x# as

#v=v_0+Kx#

(1) The acceleration #a# is given as follows

#a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v=v\frac{dv}{dx}#

#\therefore a=(v_0+Kx)\frac{d}{dx}(v_0+Kx)#

#a=(v_0+Kx)K#

#=K(v_0+Kx)#

(2). We know that the velocity #v# is given as

#v=\frac{dx}{dt}#

#dt=\frac{dx}{v}#

#\therefore dt=\frac{dx}{v_0+Kx}#

#\int dt=\int \frac{dx}{v_0+Kx}#

#\int_0^t dt=\int_{0}^x \frac{dx}{v_0+Kx}#

#t=[\frac{\ln(v_0+Kx)}{K}]_{0}^x #

#t=1/K\ln(\frac{v_0+Kx}{v_0})#

#t=1/K\ln(\frac{v}{v_0})\quad (\because v_0+Kx=v)#

(3) Slope of velocity #v# vs displacement #x# curve is given as

#\frac{dv}{dx}=\frac{d}{dx}(v_0+Kx)=K#