Any physics buffs care to help me out?

<p>Hey guys. Well I was absent a few dayss and now I am totally confused in physics. We’re doing forces/newtons laws and I totally dont understand it. If its not a bother, could aanyone please explain to me how to do this problems? Like how you would approach them etc.</p>

<li><p>After a day of testing race cars you decide to take your own 1550-kg car onto the test track. While moving down the track at 10.0 m/s you uniformly accelerate to 30.0 m/s in 10.0 s. What is the average net force that the track has applied to the car during the 10.0s interval?</p></li>
<li><p>A 65-kg swimmer jumps off a 10.0m tower.
a. Find the swimmers velocity on hitting the tower
b.The swimmer comes to a stop 2.0m below the surface. Find the net force exerted by the water.</p></li>
</ol>

<p>Im not trying to get you guys to do my hw, you dont need to give the answers just maybe explain how to do these questions because I totally dont understand forces. :(</p>

<p>Draw a free body diagram. Split anything that is not horizontal or vertical into components. List all your x and y forces. Solve for the missing force. Sorry I can't elaborate, I have to go. If someone doesn't do it before me, I'll get back and explain around 8.</p>

<p>Ok for number 1:
F=ma, you know mass, you need to find acceleration
acceleration= change in velocity/change in time
you sped up 20m/s in 10 seconds, so a=20/10=2
2*1550=your answer, in Newtons</p>

<p>number 2
velocity=at+initial velocity
a=-9.8, initial velocity is 0
you still have 2 unknowns, so you go to position function
P=.5at^2+initialvelocity<em>t+initalposition
so
0=.5(-9.8)t^2+10
-20=-9.8t^2
20/9.8square rooted=t
plug this number into t into the equation -9.8</em>t and that is your answer
for the next part, just write another position and velocity equation and solve them the equations will be:
-2=.5at^2+theanswerforthelastpart*t
and
0=at+answerforlastpart
solve for a, multiply by mass</p>

<p>We never learned position function I dont believe. Is there another way to do that?</p>

<p>not that I know of, you need those constant acceleration formulas, at least that's how I work these type of problems out:
xf=1/2atsquared+vot+xo
change in x= 1/2atsquared+vot
vf=at+vo
vfsquared=vosquared+2Achange in x</p>

<p>they're pretty simple you can basically derive all of them from the formula for acceleration</p>

<p>Obviously, it's ten feet away.</p>