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MakcuM [25]
3 years ago
6

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

(t) is its acceleration. A particle moves along the x-axis at a velocity of v(t) = 5/√t, t > 0. At time t = 1, its position is x = 12. Find the acceleration and position functions for the particle.
Physics
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer:

We know that the acceleration of the particle is defined as

a(t)=\frac{dv}{dt}

Since it is given that

v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}

Now by definition of velocity we have

v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt

Integrating on both sides we get

v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt

Applying values we get

v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2

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