with this Acceleration Problem

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The discussion revolves around a scenario where a driver follows a fire truck at a distance of 32 meters while both are traveling at 110 km/h. After the truck begins emergency braking at 5 m/s², the driver loses attention for 2 seconds, leading to confusion about the distance between the two vehicles when attention is regained. Initial calculations incorrectly estimated the separation distance, but after applying the correct formulas, the final separation was determined to be 22 meters. The second part of the problem involves the driver beginning to brake after an additional 0.4 seconds, but the initial speed calculation upon impact was incorrect. The conversation emphasizes the importance of using the right formulas for distance and velocity in physics problems.
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You are driving your car while following a fire truck by 32 m. Both your car and the truck are traveling at 110 km/h. Your attention diverts from the truck for 2.0 s At the beginning of that 2.0 s, the truck begins emergency braking at 5 m/s2.

1)What is the separation between the two cars when your attention finally returns?
I converted the velocitys to m/s and got that 110 km/h was 275/9 m/s. Then I tried to find the displacement after two seconds and got that the car traveled approximately 61 m and the truck 73.6m. Taking into account the 32m they were orginally separated by, i got that the answer was 45 m.

This is wrong and I don't know what I did.Please help if possible.
 
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What does "Your attention diverts your attention from the truck..." mean? And where did the police officer come from?
 
e(ho0n3 said:
What does "Your attention diverts your attention from the truck..." mean? And where did the police officer come from?
lol...
 
Could you clean up the wording on that a little bit? Go ahead and state exactly what the book says, please.

It would seem to me you didn't break while the truck did, so the truck should have traveled less distance than you. That's if the police officer is driving the truck.

The distance you travel in the car seems accurate by what I get. Right now, I'm thinking for the truck, you take

Distance = V0(t) + 1/2(a)(t^2), because a is negative, it will actually be
V0(t) - (1/2)(a)(t^2) for the distance the truck travelled.

Could you show me what you did for the truck please?

Next, you take the distance you traveled - the distance he travelled. Then the distance between you two is the original distance - the difference. Let me know what you get.
 
Thank You

I used the formula you gave me for the truck and got that it traveled 51 m. After doing the algebra I got the final answer to be 22m and it was correct. Thank you.

Now the second part states:
Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the truck. I used the formula
v=v0 + at and got the answer to be 1.11 km/h and it was wrong: I used v0= 275/9 m/s, a= -5/ms2, and t=2.4s and then converted to proper units.
PS: Sorry for the wording, that's what happens when you type in a rush
 
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I kinda figured what you were getting at. Glad I could help.
 
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