# Homework Help: Uniform Motion

1. Jun 4, 2013

### Nirupt

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

Cletus is travelling at 30.0 m/s as he passes a trooper hidden behind a billboard. Four seconds later, the trooper begins his chase for Cletus with a constant acceleration of 2.00 m/s2. How long does it take the trooper to intercept Cletus?

2. Relevant equations

x = vAVG.t = (v 0 + v)/2*t {Acceleration-Not-Known, or ank}

v =v0 +at {Displacement-Not-Known, or xnk}

v 2 = v0 2 + 2ax {Time - Not - Known, or tnk}

x =v0t + ½*at 2 {Final Velocity - Not - Known, or vnk}

3. The attempt at a solution

I'm kind of stuck..

I know that d1 = d2 because they will eventually meet. I know that the first part the velocity is constant at 30 m/s so acceleration is 0. The trooper has an initial velocity of 0 until the 4 second mark which the acceleration is 2 m/s^2, so 1) since the equation starts at 4 seconds would the initial velocity not be 0? and is t(4) really time = 0? in which case I am finding time, and I know I have to set up the 2 cars set equal to each other, and the Δχ displacement is not known but it is known that they will eventually meet at the same point.

Just want to know if my thought process is correct or am I digging myself in a hole. I doubt this is a hard problem

2. Jun 4, 2013

### Staff: Mentor

okay so you know d1=d2 and d1=vcletus * t

then what is d2 = ? what equation would you select from your list? (hint look at the last one x= ...)

3. Jun 4, 2013

### Nirupt

x =v0t + ½*at 2 but V0 = 0 so it would just be x = .5*at, but that would be 2 unknowns (time, and displacement)

4. Jun 4, 2013

### Staff: Mentor

didn't you say that d1=d2 so why aren't you equating the two equations and solving for t?

5. Jun 4, 2013

### CAF123

Would you not need to make an adjustment to t in the expression for d2 to take into consideration the fact that the trooper begins his chase 4 seconds later? And then solve for t?

6. Jun 4, 2013

quite right/

7. Jun 4, 2013

### Nirupt

v0t + ½*at^2 = v0t + ½*a(t+5)^2
0 = + t^2 - 30 m/s*t + 25

This might be crazy but this does not work I know that much lol I set the last equation but added +5 because the trooper had a 5 second loss.. I just feel like I'm missing something so simple. I'm not getting any of these answers.

22.3 s

15.4 s

49.1 s

33.6 s

8. Jun 4, 2013

### Staff: Mentor

why do you have the 1/2 at^2 on the cletus side. he's not accelerating.

at t0 cletus passes the trooper

at t4 the trooper begins the pursuit. where is cletus? he's 4 * 30 m/s away = 120 meters ahead

at some time t, the trooper catches cletus. cletus has traveled 120 + 30 * (t) and the trooper has traveled 1/2 a (t)^2

solve for t. this t is time since the officer started the pursuit so you need to add 4 secs to it.

now do you see how to solve it?

there is probably a more formal way to solve this.

9. Jun 5, 2013

### Nirupt

I see.. I had no idea about the simple step of multiplying 4 seconds, blew my mind haha.

120 + 30t = t^2
0 = T62 - 30t - 120

x=33.57417562100671 x=−3.5741756210067095

10. Jun 5, 2013

### CAF123

Would you need to add on the 4 secs here as well? Doesn't the 30(4) take that into account already?

I think you can do it slightly quicker by saying: $$30t = \frac{1}{2}a(t-4)^2,$$ but solving gives t = 37.6s which is 4s out from the proposed answer.

Any thoughts on why these are not the same?

11. Jun 5, 2013

### Nirupt

Well since the trooper was 4 seconds late, you would still have to subtract 4 seconds, so the answer would be 33.6s because it's (t-4)

12. Jun 5, 2013

### CAF123

I am not as yet convinced. Writing $30t = \frac{1}{2}at^2$ would mean they both left at the same time. Clearly they are at the same distance when t=0 and there exists another time t=2(30)/a.

However, by writing the troopers displacement as 1/2 a (t-4)2, I have defined the origin where they pass and shifted the displacement graph of the trooper along the t axis in the positive direction.

13. Jun 5, 2013

### Staff: Mentor

you have to decide what time you're looking for:

1) is it when the trooper sees cletus passby
2) or when the trooper starts his pursuit 4 secs later

1/2 a t^2 = v * (t + 4) for the first choice and remembering to add the 4 seconds to the answer

or

1/2 a t^2 = 120 + v * t for the second choice

14. Jun 5, 2013

### CAF123

Assume the origin is where cletus and the trooper pass. The displacement of cletus is a straight line x = 30t which starts at (x=0,t=0). The displacement of the trooper is x = 1/2 (2) (t-4)2 = (t-4)2, which is x = t2 shifted 4s along the t axis. (i.e the displacement of the trooper from the origin at t=4 is zero and then he starts moving).

I cannot see the fault with this reasoning but I am out by 4s in the result.

15. Jun 5, 2013

### Staff: Mentor

because your equation says the trooper is accelerating between t0 and t4

16. Jun 6, 2013

### CAF123

If I were to sketch the graphs and truncate x = (t-4)2 for t < 4 in the picture (so that it starts at x = 0 at t=4) then, while the graphs makes sense, (I.e at t = 4, cletus is at some x and the trooper is at x =0 in [0,4]), I still get 37.6s.

http://m.wolframalpha.com/input/?i=solve+30x+=+(x-4)^2&x=8&y=1

17. Jun 14, 2013

### Staff: Mentor

There's nothing wrong with your reasoning or method, just your interpretation of what is being asked for. You measure time from the moment they pass. But the question asks "How long does it take the trooper to intercept Cletus?". Presumably they mean as measured from the moment the trooper takes off. So just subtract 4 seconds.

18. Jun 14, 2013

### CAF123

Yes, it was the time from where the trooper took off that I overlooked. Thanks.