Question

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

Answer 1

`v = \sqrt{v_0^2 - (2k)/(3)(s^3 - s_0^3)}`

v = 9.67 m/s

Step-by-step explanation:

As we know that acceleration is rate of change in velocity

so it is defined as

`a = (dv)/(dt)`

`a = v(dv)/(ds)`

here we know that

`a = - ks^2 = v(dv)/(ds)`

now we have

`vdv = - ks^2ds`

integrate both sides we have

`\int vdv = -k \int s^2ds`

`(v^2)/(2) - (v_0^2)/(2) = -k((s^3)/(3) - (s_0^3)/(3))`

`v^2 = v_0^2 - (2k)/(3)(s^3 - s_0^3)`

here we know that

`v_0 = 10 m/s`

`s_0 = 3 m`

`v^2 = 10^2 - (2(0.10))/(3)(5^3 - 3^3)`

`v = 9.67 m/s`