A certain particle moves along the x-axis shown in the eqn #t=2x^2-5x+3# where x is measured in cm and t in seconds. Initially, the particle is 1.5cm to the right of the origin. How do you prove the velocity v cm/s is given by #v=1/(4x-5)#?

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Answer 1 · 2018-01-28T12:10:03

Given, #t=2x^2-5x+3#

Now,we can differentiate this w.r.t #x#

So, # (dt)/(dx) = 4x -5# ...1

Now, velocity = # (dx)/(dt)# i.e change in rate of displacement

So, we can write,equation of velocity is #v=(dx)/(dt) = 1/(4x-5)# (from 1)

Answer 2 · 2018-01-30T09:56:02

Given the expression

#t=2x^2−5x+3#

To find velocity we need to calculate #dotx#
Differentiating both sides with respect to #t# we get

#d/dt(t=2x^2−5x+3)#
#=>1=4x\ dotx−5\ dotx#
#=>(4x-5)\ dotx=1#
#=>dotx=1/(4x-5)#