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The acceleration of a particle is given by a = −ks2 , where a is in meters per second squared, k is a constant, and s is in meters. Determine the velocity of the particle as a function of its position s. Evaluate your expression for s = 5 m if k = 0.1 m−1 s−2 and the initial conditions at time t = 0 are s0 = 3 m and v0 = 10 m /s

Respuesta :

Answer:

[tex]v = \sqrt{v_0^2 - \frac{2k}{3}(s^3 - s_0^3)}[/tex]

v = 9.67 m/s

Explanation:

As we know that acceleration is rate of change in velocity

so it is defined as

[tex]a = \frac{dv}{dt}[/tex]

[tex]a = v\frac{dv}{ds}[/tex]

here we know that

[tex]a = - ks^2 = v\frac{dv}{ds}[/tex]

now we have

[tex]vdv = - ks^2ds[/tex]

integrate both sides we have

[tex]\int vdv = -k \int s^2ds[/tex]

[tex]\frac{v^2}{2} - \frac{v_0^2}{2} = -k(\frac{s^3}{3} - \frac{s_0^3}{3})[/tex]

[tex]v^2 = v_0^2 - \frac{2k}{3}(s^3 - s_0^3)[/tex]

here we know that

[tex]v_0 = 10 m/s[/tex]

[tex]s_0 = 3 m[/tex]

[tex]v^2 = 10^2 - \frac{2(0.10)}{3}(5^3 - 3^3)[/tex]

[tex]v = 9.67 m/s[/tex]

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