A: How far does a car traveling at 90.0 km/h take to stop?

[cherryrocket]

7.Q: "Nancy was riding in a helicopter which was rising at 4.0 m/s when she dropped her camera out. After 2.0 s how far apart were the camera and the helicopter?"
A: This isn't an acceleration question is it? Just velocity?
d=vt = (4.0m/s)(2.0s) = 8.0m

8.Q: "A certain car tire can accelerate a car to a stop at a maximum rate of -11.2 m/s^2".
A driver going at 90.0 km/h sees a hole 30.0m ahead and hits the brakes. How far does it take the car to stop? Does the car hit the hole? Assume a reaction time of zero seconds."
A:
a = -11.2 m/s^2
v = 90.0 km/h = 24.993 m/s = 25.0 m/s

t=v/a = 25.0m/s / -11.2 m/s^2 = 36.2s

d=vt = 25.0m/s / (36.2s) = 0.691 m

"The car stops 0.691 m later, so no it does not hit the hole." Is this correct?

 

[learningphysics]

For number 7, yes it is an acceleration problem...

You got the distance the helicopter traveled in 2.0s. So the helicopter has a vertical displacement of 8.0m from the place where the camera is dropped...

But what is the camera's displacement? The camera also moves.

For number 8,

you seem to be doing something wrong with your calculations... 25/11.2 is not 36.2

And why are you dividing instead of multiplying when you calculate d = vt ?

 

[cherryrocket]

Nancy's dropped camera REvised

okay, so first I went:
d=vt =4m/s(2s)
= 8m.
So 8m is the distance that the helicopter rised in 2 secs, from the time the camera was dropped.

Then I look for the final velocity of the camera, because I will need it in a formula.
Vf = 4m/s + (-9.81m/s^2 X 2sec)
Vf = 4 m/s + (-19.62 m/s)
Vf = -15.62 m/s a formula for distance:
d = (Vf - Vi)/2a
= (-15.62 m/s - 4m/s) / 2(-9.81m/s^2)
= -19.62 m.s^2/-19.62m/s^s
= 1 m

So the Camera went down 1 m from where Nancy dropped it. Thus, the helicopter and the camera are 9 m apart.
Is this right?
I think negatives and positives (if I didn't get them right) are going to screw up my answer big time.

 

[cherryrocket]

Someone please answer my question. I really need it... it is the last question on my assignment and I need to understand it... thnks

 

[learningphysics]

In this part:

"d = (Vf - Vi)/2a
= (-15.62 m/s - 4m/s) / 2(-9.81m/s^2)
= -19.62 m.s^2/-19.62m/s^s
= 1 m"

You need d = (Vf^2 - Vi^2)/2a

which gives d = -11.62

The minus sign indicates the camera went down... if there was no minus sign, then the camera went up (which could definitely have been a possibility if they chose a different time instead of 2s).

So the answer is 8 + 11.62 = 19.62m

Another way to look at it... the position of the helicopter is 1m (taking the start point as 0m)

The position of the camera is -11.62m

The displacement from the helicopter to the camera is position of helicopter - position of camera

8 - (-11.62) = 19.62m (that way, the minus signs are taken care of)

You could have also calculated the position of the camera with:

d = v1*t + (1/2)at^2

d = 4*2 + (1/2)(-9.81)*2^2 = -11.62m (same answer as before).

anyway, your answer is 19.62m.

 

[chocokat]

So the Camera went down 1 m from where Nancy dropped it. Thus, the helicopter and the camera are 9 m apart.
Is this right?
I think negatives and positives (if I didn't get them right) are going to screw up my answer big time.

For this question, another way to 'verify' your answer is to think about it. The amount the helicopter has risen is correct, and it makes sense, however the idea that a camera would have only fallen 1 m in two seconds, hmmm, just doesn't ring true. I know you can't often use this type of thinking, but for this particular problem it is useful.

 

[cherryrocket]

Thanks everyone for your help!