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
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
Well, hello there, confused human! Don't worry, I'll help you out with some humor-infused math.
Let's start with the equation d(t) = 1/2 * gt^2, where g is 10 m/s^2, as you mentioned. We're interested in finding the distance covered in time t = n seconds.
To find the distance covered in t = n seconds, we simply substitute t with n in the equation:
d(n) = 1/2 * g * n^2.
Voila! That's the distance covered in n seconds.
Now, let's move on to the distance covered in t = n - 1 seconds. Here's where we'll have some fun:
To find the distance covered in t = n - 1 seconds, we substitute t with (n - 1) in the equation:
d(n-1) = 1/2 * g * (n - 1)^2.
Now, if you want to simplify this expression further, you can expand and simplify the squared term, but you might end up with a more complex equation... and that wouldn't be much fun, now would it?
Hope this answer puts a smile on your face. Keep on falling (not literally), dear human!
To find the distance a falling body covers in time t = n seconds, we can substitute n for t in the equation d(t) = 1/2 gt^2.
So, the distance covered in time t = n seconds is: d(n) = 1/2 gn^2.
Now, let's find the distance covered in time t = n - 1 seconds. We can substitute n - 1 for t in the same equation.
The distance covered in time t = n - 1 seconds is: d(n - 1) = 1/2 g(n - 1)^2.
Simplifying both equations:
For d(n):
d(n) = 1/2 gn^2
For d(n - 1):
d(n - 1) = 1/2 g(n^2 - 2n + 1)
= 1/2 gn^2 - gn + 1/2 g
Therefore, in terms of n, the distance covered in t = n seconds is 1/2 gn^2, and the distance covered in t = n - 1 seconds is 1/2 gn^2 - gn + 1/2 g.
To find the distance a falling body covers in time t = n seconds, we can substitute n into the equation d(t)=1/2 gt^2, where g is the acceleration due to gravity (10 m/s²).
So, substitute t = n into the equation:
d(n) = 1/2 g(n^2)
This gives you the distance covered by the falling body in n seconds.
Now, to find the distance covered in t = n - 1 seconds, we can substitute t = n - 1 into the equation:
d(n-1) = 1/2 g(n-1)^2
This gives you the distance covered by the falling body in (n-1) seconds.
Let's simplify these expressions to further comprehend the results:
1. Distance covered in t = n seconds:
d(n) = 1/2 g(n^2)
= 1/2 * 10 * (n^2)
= 5n^2
Therefore, the distance covered by the falling body in n seconds is given by 5n^2.
2. Distance covered in t = n - 1 seconds:
d(n-1) = 1/2 g(n-1)^2
= 1/2 * 10 * (n-1)^2
= 5(n-1)^2
Therefore, the distance covered by the falling body in (n-1) seconds is given by 5(n-1)^2.
To summarize:
- The distance covered in n seconds is 5n^2.
- The distance covered in (n-1) seconds is 5(n-1)^2.
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