The speed s of an object dropped from rest varies with the distance y it has fallen according to the equation s=2gy−−−√ , where g is the acceleration due to gravity. What is the instantaneous rate of change of the speed with respect to distance? (1 point) Responses dsdy=12gy−−−√ d s d y = 1 2 g y dsdy=1gy−−√ d s d y = 1 g y dsdy=g2y−−√ d s d y = g 2 y dsdy=gy−−√ d s d y = g y dsdy=2gy−−√
The correct response is: d s d y = g 2 y
To find the instantaneous rate of change of the speed with respect to distance, we need to differentiate the equation s=2gy^0.5 with respect to y.
Let's differentiate both sides with respect to y:
ds/dy = d/dy (2gy^0.5)
Using the power rule of differentiation, we can bring down the exponent and multiply it by the coefficient:
ds/dy = 2g * 0.5y^(-0.5)
Simplifying further:
ds/dy = g * y^(-0.5)
Thus, the instantaneous rate of change of speed with respect to distance is ds/dy = g * y^(-0.5).
To find the instantaneous rate of change of the speed with respect to distance, we need to calculate the derivative of the equation s=2gy^0.5 with respect to y.
The derivative of s with respect to y can be found using the power rule of differentiation. The power rule states that if we have an expression of the form f(x) = ax^n, then the derivative of f(x) with respect to x is given by f'(x) = nax^(n-1).
In our case, s = 2gy^0.5. Using the power rule, the derivative of s with respect to y is:
ds/dy = (0.5)(2g)(y^(-0.5))
ds/dy = g/y^0.5
So, the instantaneous rate of change of the speed with respect to distance, ds/dy, is equal to g/y^0.5. Therefore, the correct response is:
ds/dy = g/y^0.5