The velocity of a particle moving in the positive direction of X-axis varies as v = a root x where a is positive constant. Assuming that at the moment t = 0 ,the particle was located at x = 0 the value of time dependence of the velocity and acceleration of the particle
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v = a √x
x = 0 at t = 0
v =dx/dt = a√x
dx/√x = a dt
2 √x = a t + c
x = 0 at t = 0 c = 0
x = a²t²/4 dx/dt but differentiate with respect to time
v=a²t/2differentiate with respect to time
acc^n = a² / 2
we can do this equation by s=ut+1/2at²
praneeth1803:
i dont have an option a/2√x
options are a) 2/2a² b) a²/2 c )2/a²
Answered by
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given
v = a √x
x = 0 when t = 0
v = dx / dt = a √x
=> dx/√x = a dt
integrate on both sides
=> 2 √x = a t + c
x = 0 at t = 0 => c = 0
=> x = a²t²/4 differentiate now: wrt t
=> v = a² t /2 differentiate now wrt t
=> acceleration = a² / 2
v = a √x
x = 0 when t = 0
v = dx / dt = a √x
=> dx/√x = a dt
integrate on both sides
=> 2 √x = a t + c
x = 0 at t = 0 => c = 0
=> x = a²t²/4 differentiate now: wrt t
=> v = a² t /2 differentiate now wrt t
=> acceleration = a² / 2
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Please tell me the value of time dependence of the velocity
a square t / 2
acceleration is constant a square /2
how u got a square t /2
differentiation: dx/dt = v......differentiation of t^2 gives 2 t.
if you donot know differentiation, yet.. then do this:
x1 = t^2.... x2 = x1 + dx ;;; t2 = t1 + dt;;; x+dx = (t+dt)^2 ;;; x2 - x1 = (t+dt)^2 - t^2
x2 - x1 = 2 t dt + dt^2.;;; ( x2 - x1) / (t2 - t1) = 2 t neglecting dt which tends to 0.
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