# Car and satillite problem

1. Sep 30, 2007

### rgitelman

1. The problem statement, all variables and given/known data

1. Starting from rest, a car travels 1350m in one minute. It accelerated at 1m/s2 until it reached its cruising speed. Then it drove the remaining distance at constant velocity. What was its cruising speed?

2. A satellite is in a circular orbit 600km above the Earth’s surface. The acceleration of gravity is 8.21m/s2 at this altitude. The radius of the Earth 6400km. Determine the speed of the satellite and the time to complete of orbit around earth.

2. Relevant equations
problem 1:
d=1350m
t=60sec
a=1m/s2

a=v/t
v=at2
v=d/t

t1 t2
|___________.______________________> time x
d1 d2

d1+d2=1350m t1+t2=60 secs

d1=a(t1)squared d2=v(t)sqaured

problem 2:
r=600km+6400km=7000km=7,000,000m
a=9.81m/(s)squared

a=(v)squared/r
v=d/t = pie*diameter/t

3. The attempt at a solution

problem 1:
i derived these two formulas
d=a(t1)squared+v(t)squared
d=a(t1)squared+v(t-t1)

i don't know where to go from there because i have 2 unknown variables, v and t1.

problem 2:
(v)squared/7,000,000m=8.21m/(s)squared

v=7580.897045m/s

7580.897045m/s=pie*(7,000,000*2)/t

t=5801.727274s

i got the answer for speed and time, but i'm not sure if i did it right.

2. Sep 30, 2007

### JoAuSc

problem 1:

d = a*t1^2 + v*(t-t1)

is correct but you're forgetting that the cruise velocity v is also constrained by being the result of acceleration for t1 time:

v = a*t1

so you can substitute for v into the d equation and you end up with a quadratic equation with one unknown.

problem 2:

Your answers seem to be correct. Btw, your math would look much more clear if you did something like "v^2" to mean "v squared".