Question

The total distance a freely falling body covers in time, t, is given by the
equation d(t)=1/2 gt2
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.
Show, in terms of n, the distance a falling body covers in t = n - 1 seconds
of movement...I don't even know where to start, any help would be appreciated. thank you

Answer 1

so d(t) = 5t^2 , (should really be d(t) = -5t^2 )

so if t=n
d(n) = 5 n^2
n^2 = d(n) /5
n = √(d(n)/5)

You might also want to replace d(t) with h for easier typing and reading
that is, n = √(h/5)

b) replace t with n-1, repeat my steps

Answer 2

d = 5t^2
d(n) = 5n^2
d(n-1) = 5(n-1)^2 = 5n^2 - 10n + 5

just FYI, the distance fallen in the nth second is thus
d(n)-d(n-1) = 10n-5