You said it was a trick question, so I counted all the "questions" in the page. I edited my post because I thought we were only allowed to answer one question. Oh and BTW, there were a lot more than 9, lol.
INCORRECT INCORRECT and you edited your post. The purpose of a trick question is to trick you. You are incorrect in your assumptions. INCORRECT There are some major hints and helpers in these guys' answers.
INCORRECT If you have read the books and are not just googling this stuff, think in a broader perspective.
Assuming that the officer is standing still (so 0 km/h), it will take the officer 17.5 seconds to reach the speeder (so the speeder and the officer will be at the same point) 140/8 = 17.5 I'm probably wrong cause I don't think the question would be that simple xD. But worth a guess to gain some information.
so the cops acceleration from rest is 2.22222222 m / s^2 police initial velocity = 0 m/s the car's velocity is 38.888888... m/s so let the instant the car passes by the cop be our starting time... and our zero displacement point equation for cop is x=1/2(a)(t)^2 equation for car is x=vt so lets make them equal to each other vt=1/2(a)(t)^2 38.8888t=1/2(2.22222)(t)^2 this will give us the poitns at which the two cars are at the same displacement.... we have a 0 value (when the cars started cause we SET that to 0 displacement point.) 38.8888t=1.111111(t)^2 so we have a simple quadratic divide by t 38.8888=1.1111(t) approximately 35 seconds assuming that the cop catches the person once the cop catches up to the car.
That is...CORRECT Congratulations, credits coming your way. Signatures will still be a few days but you will be given your choice soon.
Ask some science related questions. -_- And don't say there was already a Physics question up there, because that's more math than science.
free bump come on guys answer these they just require a little thinking... and look at the prizes he's giving for just that