Question

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?

A) ds/dy = √(2gy)
B) ds/dy = 2√(g/y)
C) ds/dy = g/√(2gy)
D) ds/dy = g/y
E) ds/dy = 2g/√y

Answer 1

The instantaneous rate of change of speed with respect to distance for an object in free fall, described with the equation s = √(2gy), is found by differentiating s with respect to y, resulting in the derivative ds/dy = g/√(2gy).

Step-by-step explanation:

The student is asking about the instantaneous rate of change of the speed with respect to distance for an object in free fall, which can be calculated using calculus. Given the equation for speed s is equal to the square root of 2 times the acceleration due to gravity g times the distance y (s = √(2gy)), we want to find the derivative ds/dy.

To find the instantaneous rate of change of the speed with respect to distance, we differentiate s with respect to y. Applying the chain rule:

ds/dy = d/dy(√(2gy)) = (1/2)(2g)^{-1/2} * (2g) * dy/dy = g/(√(2gy))

The correct answer is therefore C) ds/dy = g/√(2gy).