If a point moves in straight line in such a manner that its acceleration is proportional to its speed (a ∝ v) , what is the relation between distance covered and speed?

a) #x prop v#
b) #x prop v^2#
c) #x prop v^3#
d) #x prop sqrtv#

a) #x prop v#
b) #x prop v^2#
c) #x prop v^3#
d) #x prop sqrtv#

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A08 Share
Dec 29, 2017

Answer:

I get (a)

Explanation:

Given that acceleration #a prop v#, speed

From definition of acceleration and velocity
#=>(dv)/(dt)propdx/dt#
Integrating both sides with time #dt#, for LHS we get

#int(dv)/(dt)dt=v+c#
where #c# is a constant of integration and can be found from initial conditions.

For RHS we get

#intdx/dt dt=x+c_1#
where #c_1# is a constant of integration and can be found from initial conditions.

Assuming that initially at #t=0#, acceleration, velocity and displacement all are #=0#. Both #cand c_1# are#=0#. We get

#vpropx#
#=>xpropv#

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