Suppose that the velocity of a falling body is v = k/s (k a constant) at the instant the body has fallen s meters from its starting point. Find the body's acceleration as a function of s.
v = Vi + a t = 0 + a t
= a t = k/s
so
t = k/(a s)
s = 0 + . 5 a t^2
= .5 a t^2
s = .5 a k^2/(a^2 s^2)
s^3 a = .5 k^2
a = .5 k^2/s^3
To find the body's acceleration as a function of s, we can start by differentiating the given velocity equation with respect to time (t).
Given:
Velocity (v) = k/s
Now, we need to find the acceleration (a), which is the derivative of velocity with respect to time.
a = dv/dt
Let's differentiate the equation v = k/s with respect to t:
a = d(k/s)/dt
To differentiate k/s with respect to t, we need to use the chain rule.
Chain Rule: d(u/v)/dt = (v * du/dt - u * dv/dt) / (v^2)
For k/s, u = k and v = s.
Differential of k with respect to t, du/dt = 0 (since k is a constant)
Differentiating s with respect to t, ds/dt = velocity (v)
Now, we can substitute these values into the chain rule formula:
a = (s * 0 - k * v) / (s^2)
Simplifying this equation:
a = -k * v / s^2
Therefore, the body's acceleration as a function of s is:
a(s) = -k * v / s^2
To find the body's acceleration as a function of its position, we need to differentiate the velocity function with respect to position. In this case, the velocity function is given as v = k/s, where k is a constant.
Differentiating v with respect to s, we apply the quotient rule for differentiation:
dv/ds = d(k/s)/ds
Using the quotient rule, the derivative of k/s with respect to s is:
= (s * d(k)/ds - k * d(s)/ds) / s^2
Since k is a constant, d(k)/ds = 0. And d(s)/ds = 1.
Therefore, the expression simplifies to:
dv/ds = (-k) / s^2
This is the derivative of the velocity function with respect to position, which represents the body's acceleration as a function of its position. So, the body's acceleration as a function of s is given by:
a(s) = -k / s^2