Question

The total distance a freely falling body covers in time, t, is given by the equation d(t) = 1/2 gt^2, where g is constant at 10 m/s². Show, in terms of n, the distance a falling body covers in t = n seconds of movement.

a) d(n) = 5n^2
b) d(n) = 10n^2
c) d(n) = 2.5n^2
d) d(n) = 20n^2

Answer 1

The total distance a falling body covers in t = n seconds can be found using the equation d(n) = 1/2 * 10 * n^2, which simplifies to d(n) = 5n^2.

Step-by-step explanation:

The total distance a freely falling body covers in time, t, is given by the equation d(t) = 1/2 gt^2, where g is constant at 10 m/s². To find the distance a falling body covers in t = n seconds of movement, we substitute n for t in the equation:

d(n) = 1/2 g(n)^2

Simplifying the equation, we get:

d(n) = 1/2 * 10 * n^2

d(n) = 5n^2

Therefore, the correct answer is d(n) = 5n^2.