quite the mind boggling question

[Krz]

"An unmarked police car, traveling a constant 90km/h, is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the police man steps on the accelerator. If the police car's acceleration is 2.00 m/s^2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at a constant speed)?"

 

blew my mind away

[nanogrinder]

i got 14.8

 

first i took 90km/h and 140km/h then converted them to m/s which comes out to be 25m/s and 38.9m/s respectively. we will call the car 25,x and car 38.9, z

 

after this i use the equation

 

d=Vot+.5at²

 

where d=distance, Vo=original velocity, a=acceleration, t=time

 

the first second for car x travels 25m, using d=25(1)+.5(0)(1²) the last part is zeroed out because there is no acceleration. car z travels 38.9m, d=38.9(1). we need to now make two equations involving car x and z that are set equal to each other,but we must also add in those other two answers in them.

 

since they are going to be at an equal distance in a certain amount of time, we can do this

 

38.9t+38.9= 25+25t+.5(2)t² (btw we had to add the other values 38.9 and 25 into the equation) now get everything to one side -t²+13.9t+13.9=0

now just put the numbers into the quadratic formula then volia 14.8s

 

i hoped that help explain, its hard to get these kinds of points across on the net.

[swansont]

The point of the HW forum is not to do the problem for the poster, but to help him/her solve the problem and learn how to do it.

[Krz]

haha its cool, i can learn from seeing what he did, although his answer was off by 1, i dont know :S

[nanogrinder]

The point of the HW forum is not to do the problem for the poster, but to help him/her solve the problem and learn how to do it.

 

sorry about that.

 

p.s. krz i forgot to add in the second it took for the officer to react.

[Krz]

cool cool, thanks

[ishanpup]

yea i got 14.8 seconds as well, but i originally simplified the problem using relativity. the speeder can be assumed to be going 50km/h while the cop is stationary. 50km/h = 13.89 m/s. 

I divided the problem into 3 parts: 

1) after 1 second

2) when cop speed = 50km/h (cop and speeder are going at same speed) 

3) when cop finally catches up to him

so when we do the math at the end, we get t1 + t2 +t3 = t(total) ], which is our answer.

(FOR THIS PROBLEM I WILL BE USING M/S UNTIL THE VERY END, SO I WONT WRITE IT ALL THE TIME) (v is implied in m/s unless otherwise stated) 

1.)after one second

cop: no displacement because stationary, v_i = vf = 0, t = 1

speeder: displacement = vt; 13.89m = (13.89m/s)(1s) 

***

1 second has passed

 

 

2.) when cop speed = 50km/h (cop and speeder are going at same speed) 

cop: v_i = 0, v_f = 13.89, a = 2 

using UAM equation: v_f = v_i + at, 13.89 = 0 + 2t, t=6.945

displacement = [(v_i+v_f)/2](t), displacement = [(0+13.89)/2](6.945) = 48.23

speeder: v_i = 13.89, v_f = 13.89 t = 6.945 

using v = displacement/time, 13.89 = displacement/6.945, displacement = 96.47

 

cop has travelled 48.23m , speeder has travelled 96.47m, their speeds are equal

***

6.945 seconds have passed JUST in this part

the cop is (96.47-48.23) = 48.24m away from the speeder

they both have a speed of 13.89

 

3) when cop finally catches up to him

cop: displacement = v_i(t) + 1/2(a)(t²), displacement + 48.24 = 13.89t + 1/2(2)(t²)

*(the +48.24 next to displacement in this equation is to account for the distance between the cop and the speeder at this point in time. to catch the speeder, the cop will have to travel the same distance as the speeder in this section PLUS make up for the gap that was already existing (48.24m))

speeder: displacement = v_i(t) +1/2(a)(t²), displacement = 13.89(t) + 1/2(0)(t²), displacement = 13.89t

(i am aware that displacement = v(t) for a constant velocity object, but i just used a UAM equation and took the long route to create continuity) 

setting displacements equal to each other: 

13.89t + 1/2(2)(t²) - 48.24 = 13.89t

*keep in mind that "t" in this part is NOT the final answer. it is only referring to the time that it takes the cop to finally catch the criminal after the previous 2 parts have occured (1 second passed and cop accelerates until his speed = speeder speed). It may be easier to think of t in this part as t3. I didn't use t1 t2 or t3 because it can be easily mistaken as a coefficient when not in subscript)

subtracting 13.89t from both sides, and moving 48.24, and simplifying:

t²  = 48.24

t= 6.945

interesting! t3 = t2

 

So now we finally solve: 

t1 + t2 + t3 = t(total)

1 + 6.945 + 6.945 = t(total)

t(total) = 14.89 seconds

it takes 14.89 seconds for the cop to catch the speeder

 

 

 

 

 

[KJW]

On 2/28/2006 at 10:04 PM, swansont said:

The point of the HW forum is not to do the problem for the poster, but to help him/her solve the problem and learn how to do it.

I think enough time has passed that we can now do the problem.