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20 seconds after the first car passes the point (let's call it point x=0) we know the car is\[ vt=20(20)=400\] meters ahead. The position of the car after that time is then given by\[x_C=400+20t\]and the position of the police car is\[x_P=30t\]we can now represent mathematically the distance between the cars, then set it to zero and solve for t.
so are we gonna put it as: 30t = 400 + 20t
so t = 40s ?
yep, that's what I got
and then from there we can find distance right? using any equation with distance
which velocity do i use to solve for distance ?
i got the distance as 1000m :O
it doesn't matter the velocity as long as you use one equation or the other, because at time t=40 our math says they should be equal\[400+20t=400+20(40)=400+800=1200m\]\[30t=30(40)=1200m\]
So the whole point is that you're not always using the kinematics equations you can use algebra to solve
Sure, why not? If you can solve it, do so in whatever way seems simplest. That keeps down typos. Don't get me wrong, you have to use the kinematic equations, but do so flexibly. They are to be changed and combined at will!
I meant you have to *learn the kinematic eqn..
but still\[x(t)=x_i+vt\]is a kinematic eqn, just a very simple one...
can you show me quickly how to solve using a kinematic equation :O
that is the kinematic above we used it, let xc be the car and xp be the police\[x_C(t)=x_i+vt=400+20t\]\[x_P(t)=x_i+vt=0+30t=30t\]or if you want to be thorough you can derive the formula from a more standard kinematic just by noting that the acceleration is zero\[x=x_i+v_it+\frac12at^2=x_i+v_it+\frac12(0)t=x_i+vt\]but that really seems like overkill
so we can assume that the acceleration is 0m/s^2
Well, there is no acceleration given, and I don't think we can solve it if there is one, so I'm gonna say 'yes' on that.
can you check out the other one i posted please. It has something to do with mass