How Fast Was the Police Car Going When It Was 27 Meters Behind Again?

[mohabitar]

A car is speed at a constant 24 m/sec in a school zone. A police car starts from rest just as the speeder passes it and accelerates at a constant rate of 5.4 m/sec2.

Also, the cop catches the car at t=8.89 s and v=48.

How fast is the police car going when it is 27 m behind the speeding car for the second time?

So, I'm not sure if the wording here is what's confusing me, but I can't get the right answer. First of all, how can the cop car ever be behind the speeding car "for the second time"? If the car is at a constant velocity and the police car keeps accelerating, then once they meet for the first time, the cop car will always be ahead, so what does it mean for it to be behind the speeding car for the second time?

I attempted the problem using basic kinematic equations to try to first find when it is 27 m behind the car: v=48+5.4t and 27=0+48t+.5(5.4)t^2. And so I solved for t and plugged it into the first equation but that didn't get me the right answer. Maybe I'm just not understanding the problem..

 

[bp_psy]

A car is speed at a constant 24 m/sec in a school zone. A police car starts from rest just as the speeder passes it and accelerates at a constant rate of 5.4 m/sec2.

Also, the cop catches the car at t=8.89 s and v=48.

How fast is the police car going when it is 27 m behind the speeding car for the second time?

So, I'm not sure if the wording here is what's confusing me, but I can't get the right answer. First of all, how can the cop car ever be behind the speeding car "for the second time"? If the car is at a constant velocity and the police car keeps accelerating, then once they meet for the first time, the cop car will always be ahead, so what does it mean for it to be behind the speeding car for the second time?

I attempted the problem using basic kinematic equations to try to first find when it is 27 m behind the car: v=48+5.4t and 27=0+48t+.5(5.4)t^2. And so I solved for t and plugged it into the first equation but that didn't get me the right answer. Maybe I'm just not understanding the problem..

The "second time" part seems weird to me.Did you try to find the speed of the police car when is 27m behind the first car. Is that the good answer? Also if you do it your way and calculate the the speed of the police car when is 27m in front of the car. You must account for the velocity of the first car in 27=0+48t+.5(5.4)t^2..V= 48m/s is the speed of the police car with respect to the starting point but you need V relative to the first car.You can then convert again to the rest frame speed.

 

[PeterO]

A car is speed at a constant 24 m/sec in a school zone. A police car starts from rest just as the speeder passes it and accelerates at a constant rate of 5.4 m/sec2.

Also, the cop catches the car at t=8.89 s and v=48.

How fast is the police car going when it is 27 m behind the speeding car for the second time?

So, I'm not sure if the wording here is what's confusing me, but I can't get the right answer. First of all, how can the cop car ever be behind the speeding car "for the second time"? If the car is at a constant velocity and the police car keeps accelerating, then once they meet for the first time, the cop car will always be ahead, so what does it mean for it to be behind the speeding car for the second time?

I attempted the problem using basic kinematic equations to try to first find when it is 27 m behind the car: v=48+5.4t and 27=0+48t+.5(5.4)t^2. And so I solved for t and plugged it into the first equation but that didn't get me the right answer. Maybe I'm just not understanding the problem..

AT time t = 0, the cop car was level the speeding car. A very short time later the cop car will have been 1m behind the speeding car [with the speeding car opening the gap quite rapidly].

When the cop car caught the speeding motorist, they were again level with each other, naturally.
Just before he caught the speeding car, the cop car will have been 1 m behind .

That is the second time he was 1m behind the speeding car!

Presumably during this pursuit, the maximum gap from cop car to speeding motorist was more than 27m.
That means the gap first open to 27m [the first time the cop car was 27m behind the speedster]. The gap then continued to open to the maximum gap, the cop car then began to close the gap, and at some point the gap was reduced to 27m [the second time the cop car was 27m behind the speedster]. The cop car continued to close the gap and eventually caught the speedster.
That's how it works.