Question

A car traveling along the highway passes a point at 20 m/s. After 20 s, a police car passes the same point traveling at 30 m/s. how far down the road, and in what time will the police reach the first car?

Answer 1

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.

Answer 2

so are we gonna put it as:

30t = 400 + 20t

Answer 3

so t = 40s ?

Answer 4

yep, that's what I got

Answer 5

and then from there we can find distance right?

using any equation with distance

Answer 6

right

Answer 7

which velocity do i use to solve for distance ?

Answer 8

i got the distance as 1000m :O

Answer 9

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\[400+20t=400+20(40)=400+800=1200m\]\[30t=30(40)=1200m\]

Answer 10

So the whole point is that you're not always using the kinematics equations
you can use algebra to solve

Answer 11

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!

Answer 12

I meant you have to *learn the kinematic eqn..

Answer 13

but still\[x(t)=x_i+vt\]is a kinematic eqn, just a very simple one...

Answer 14

can you show me quickly how to solve using a kinematic equation :O

Answer 15

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

Answer 16

so we can assume that the acceleration is 0m/s^2

Answer 17

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.

Answer 18

can you check out the other one i posted please. It has something to do with mass