Question

The distance a freely falling object drops, starting from rest, is proportional to the square of the time it has been falling. By what factor will the distance fallen change if the time of falling is three times as long?

Short answer

Answer: The distance changes by a factor of 9.

Step-by-step solution

Understanding the relationship between distance fallen and time

The distance fallen by a freely falling object, starting from rest, is proportional to the square of the time it has been falling. This can be expressed mathematically as: d ∝ t^2 where d represents the distance fallen and t is the time it has been falling. We need to find how the distance changes if the time of falling is three times as long.

Express the relationship as an equation

To express the relationship d ∝ t^2 as an equation, we need to introduce a constant of proportionality (k): d = k * t^2

Calculate the new distance fallen

Now, let's consider a situation where the time of falling is three times as long (3t). We can then calculate the new distance fallen (d') with this increased time: d' = k * (3t)^2

Calculate the factor by which the distance fallen changes

To find the factor by which the distance changes, we need to create an equation that relates the new distance (d') to the original distance (d). We can do this by dividing the new distance formula by the original distance formula: Factor = d'/d = (k * (3t)^2) / (k * t^2)

Simplify the equation and find the factor

Now, let's simplify the equation to find the factor by which the distance changes: Factor = (k * (3t)^2) / (k * t^2) = (3^2 * t^2) / t^2 = 9 So, the distance fallen will change by a factor of 9 if the time of falling is three times as long.