|
|
|
|
|
|
|
|
|
 Posted: Sun Aug 12, 2007 8:09 pm
I didn't copy the homework question but from a friend.
I couldn't understand one word because she wrote in cursive.
A car speeding @ 25m's. A police starts from rest(i cant read this word it looks like rest to me) just as the speeder passes acceleration at a constant rate 5m'sSquared
a)when does the police catch the car?
b)How fast is the police car traveling when it catches the speeder?
|
 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sun Aug 12, 2007 8:16 pm
okay i know:
v=25m/s
a=5m/sSquared
t=?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sun Aug 12, 2007 9:39 pm
|
|
|
|
|
|
|
|
|
|
Posted: Sun Aug 12, 2007 9:58 pm
I don't suppose you recall the basic kinematic formulas?
What you want to do is to set up some basic equations:
Since the car has a constant velocity, its position at time t is given by xc = int_0^t v(k)dk = v*t, where we assume that t has units of seconds xc is in meters. This is simple distance = velocity*time.
Now, the police car's position is slightly more complicated since it is accelerating, but not much more complicated.
The distance an object travels when it has a constant acceleration is given by
xp = int_0^t v(k)dk, where v(k) = int_0^k a(s)ds + vi, where vi is the object's initial velocity.
Thus we get int_0^t (int_0^k a(s)ds+vi)dk = int_0^t (ak+vi)dk = vit+(a*t^2)/2.
In this case, it is 0, since the police car starts at rest. a in this case is 5.
So for part a, we want to solve for when xc = xp, i.e. we want to solve v*t = (a*t^2)/2, where v is the car's velocity and a is the police car's acceleration.
For part b, the velocity of the police car at time t is given by velocity = acceleration*time, i.e. vf = a*t, so you can just plug your results from part a in.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Aug 13, 2007 1:00 am
oh my gosh, you lost me.
all i know for kinematic equations are:
vf=vi+at
vfsquared=visquared+2aD
D=vit+1/2atsquared
v-speed
a-acceleration
t-time
D=distance
the equations you showing me look complicated.
i dont know what the symbols mean
xc = int_0^t v(k)dk = v*t?
xp = int_0^t v(k)dk?
v(k) = int_0^k a(s)ds + vi?
v*t = (a*t^2)/2?
where did k come from?
is int_0=initial velocity?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Aug 13, 2007 1:49 am
Hehe, sorry. I tend to think of the kinematic equations in terms of calculus, rather than on their own. Don't worry about k.
Okay, then, starting over:
Let's write D_c for the distance that the speeding car goes, and D_p for the police car.
Let's say that t = 0 is when the speeding car passes the police car.
Now, for the speeding car, the speed is constant, i.e. a is 0 for the speeding car, so we can write D_c = v*t, where v is the speed of the speeding car.
For the police car, we have D_p = vit + 1/2at^2, but in this case vi is 0, so we write D_p = 1/2at^2.
We want to find when D_c=D_p, i.e. we want to find a t such that v*t = 1/2at^2. We can cancel a factor of t from both sides to get v = 1/2at, and then divide by 1/2a to get 2v/a = t. So that is when the police car catches the speeding car.
Now, the speed of the police car at time t is given by vf = vi+at; in this case, again vi = 0, so we are left with vf = at, where t is what you got before.
Does that help?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Aug 13, 2007 3:31 am
yes, that helps big time.
thank you sooooooooooooooo much.
i think i understand now.
im 15 and am taking algebra 2.
i got kinda confused on how to approach the problem but im fine now thanks to you.
THANK YOU!!!!
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Jul 07, 2008 4:23 pm
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Fri Jul 18, 2008 2:21 pm
you could use suvat equations (kinematics)
s distance/displacement
u initial speed
v final speed
a acceleration
t time
so you know u (0), v and a so to work out t you use the equation
v=u+at
which should give you the right answer to the first part
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Reply
