a spaceship leaves the alien mother-ship and begins descending to earth. it descends 64m during the first second and 7m less during each second afterward. if the aliens invade after 10s, how far did the ship descend?

Vi = -64 m/s (velocity is down)

dv/dt = a = +7m/s^2 (acceleration is up)

h = hi + Vi t + (1/2) a t^2
h - hi = -64(10) + 3.5 (100)
= -640 + 350
= - 290
so it went down 290 meters

would there be any formulas you would have to use to get this answer for grade 11 math

h = hi + Vi t + (1/2) a t^2

usually
x = Xi + Vi t + (1/2) a t^2

In first one second it descends 64 m ,

Next one second it descends =(64-7)=57 m
Then next one second it descends = ( 57-7)=50 m,
4th second it descends =50-7=43 m
5th second it descends = 43-7=36 m.
6th second it descends = 36-7=29 m.
7th second it descends = 29- 7= 22 m
8th second it descends = 22-7=15 m
9th second it descends = 15-7=8 m.
10th second it descends = 8-7=1 m

Total distance covered by it= 64+57+50+43+36+29+22+15+8+1= 325 m

To find out how far the spaceship descended, we can use the concept of arithmetic progression. In this case, the first term is 64m, and the common difference is -7m (since the spaceship descends 7m less each second). We need to sum the terms of this arithmetic progression up to the 10th term.

One way to calculate this is by using the formula for the sum of an arithmetic progression, given by:

Sn = (n/2)(a1 + an)

Where:
- Sn is the sum of the first n terms
- n is the number of terms
- a1 is the first term
- an is the nth term

Now, let's calculate the sum of the first 10 terms:
- n = 10 (since the aliens invade after 10 seconds)
- a1 = 64m (the first term)
- an = a1 + (n-1)d, where d is the common difference
= 64 + (10-1)(-7)
= 64 + 9(-7)
= 64 - 63
= 1m

Substituting these values into the formula, we have:
Sn = (10/2)(64 + 1)
= 5(65)
= 325m

Therefore, the spaceship descended a total of 325 meters.