Given
v = sq.root of g^2 + 2gs
where g is a constant and v is velocity and s is a positive function. Show that acceleration is a constant.
To show that acceleration is a constant, we need to find the expression for acceleration and determine if it is independent of the variable 's'.
Acceleration (a) is the derivative of velocity (v) with respect to time (t). Since we don't have a direct expression for v as a function of t, we need to find the expression for v in terms of the given variables, g and s, and then differentiate it with respect to time.
Given: v = √(g^2 + 2gs)
Now, let's differentiate v with respect to t using the chain rule:
dv/dt = d(v)/d(s) * d(s)/d(t)
To calculate the first term, d(v)/d(s), we can rewrite the expression for v in terms of s:
v = √(g^2 + 2gs) = (g^2 + 2gs)^(1/2)
Now, applying the chain rule, we can calculate d(v)/d(s):
d(v)/d(s) = (1/2) * (g^2 + 2gs)^(-1/2) * d(g^2 + 2gs)/d(s)
= (1/2) * (g^2 + 2gs)^(-1/2) * d(g^2)/d(s) + d(2gs)/d(s)
= (1/2) * (g^2 + 2gs)^(-1/2) * 0 + 2g
= 2g
Now, let's calculate the second term, d(s)/d(t), which is simply the derivative of s with respect to t:
d(s)/d(t) = s' (the derivative of s with respect to t)
Finally, substituting the values back into the initial equation for dv/dt, we get:
dv/dt = 2g * s'
Since both terms on the right-hand side of the equation are constants (g is given as a constant, and s' represents the derivative of s with respect to t), we can conclude that dv/dt (acceleration) is indeed a constant.
Therefore, we have shown that acceleration is a constant based on the given equation for velocity.