The velocity of particle moving along a straight line increases according to the lawv=v_0+Kxv=v0+Kx , where K is a positive constant,then options are below?
- the acceleration of the particle is
K(v_0+Kx)K(v0+Kx)
- the particle takes a time
(1/k)*ln(v/v_0)(1k)⋅ln(vv0) 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)
- the acceleration of the particle is
K(v_0+Kx)K(v0+Kx) - the particle takes a time
(1/k)*ln(v/v_0)(1k)⋅ln(vv0) 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)
1 Answer
Jul 2, 2018
Above three options are correct
Explanation:
Given the velocity of particle as a function of displacement
(1) The acceleration
(2). We know that the velocity
(3) Slope of velocity