Solving for Average Acceleration: Slippery Road Skidding Car

[negation]

Homework Statement

Attempting to stop on a slippery road, a car moving at 80 km/h skids across the road at a 30 degree angle to its initial motion, coming to a stop in 3.9 s.

Determine the average acceleration in m/s^2, using a coordinate system with the x-axis in the direction of the car's original motion and the y-axis toward the side of the road to which the car skids.

The Attempt at a Solution

I did a whole chunk of workings but they're aren't working out to give me a sensible answer.
Could someone give me a leg up?

0ms^-1 = 22.2ms^-1 + acos30°(3.9s)
a = -6.57ms^-2

vx = 22.2ms^-1 + (-6.57ms^-1 cos30)(3.9s)
vy = (-6.57ms^-1 sin30)(3.9s)

I square root the square of vx and vy to get the resultant, then, divide |v| by t = 3.9s but it's not yield -5.7ms^-1.

What is wrong here?

 

[voko]

Why do you even need a coordinate system here? Is that a requirement for this problem?

If so, none of your equations makes sense for me. If the initial velocity is at 30 degrees to the x axis, what are its x and y components?

Then, the final velocity must have zero x and y components. Determine the components of acceleration required for that.

 

[negation]

Why do you even need a coordinate system here? Is that a requirement for this problem?

If so, none of your equations makes sense for me. If the initial velocity is at 30 degrees to the x axis, what are its x and y components?

Then, the final velocity must have zero x and y components. Determine the components of acceleration required for that.

Isn't it the acceleration magnitude that is at 30 degrees to the x-axis?

ax = |a|cos30°
ay = |a|sin30°

 

[voko]

"skids across the road at a 30 degree angle to its initial motion" is a condition on velocity.

 

[negation]

"skids across the road at a 30 degree angle to its initial motion" is a condition on velocity.

If so, then I've obtained the answer.

|a→| = Δv/Δt = [SQRT(22.2ms^-1 cos 30°)^2 + (22.2ms^-1 sin 30°)^2]/3.9s

|a→| = 5.7ms^-2
a→ = -5.7ms^-2

 

[voko]

Look what you did. (1) You found the magnitude of velocity: 22.2 m/s. (2) Then you decomposed it into the x and y components. (3) Then you found the magnitude of velocity from the components. (4) Then you divided it by time.

Were steps (2) and (3) really required?

 

[negation]

Look what you did. (1) You found the magnitude of velocity: 22.2 m/s. (2) Then you decomposed it into the x and y components. (3) Then you found the magnitude of velocity from the components. (4) Then you divided it by time.

Were steps (2) and (3) really required?

It was not-22.2ms^-1/3.9s= 5.69ms^-1 - but doing so allows me to better appreciate what is going on.

 

[voko]

Then I suggest something else.

Once you know the components of velocity, find what acceleration is required for each component to become zero in 3.9 s. This is a rephrasing of my post #2.

 

[negation]

Then I suggest something else.

Once you know the components of velocity, find what acceleration is required for each component to become zero in 3.9 s. This is a rephrasing of my post #2.

x-component of velocity = 19.22ms^-1
y-component of velocity = 11.1ms^-1

vxf = vxi +Δv = 22.2ms^-1 + 19.22ms^-1 = 41.42ms^-1
(my reasoning is based on the premise the initial velocity = 22.2ms^-1 but at the point at which p(0,0), the car skids at 30° to the x-axis with a velocity of magnitude 22.2ms^-1.

vyf = vyi + Δv = 0ms^-1 + 11.1ms^-1

|a→|x = Δvx/Δt = [vxf-vxi]/3.9s = -41.42ms^-1 / 3.9s = -10.6205ms^-2
|a→|y = Δvy/Δt = [vxf-vxi]/3.9s = -11.1ms^-1 / 3.9s = -2.846ms^-2

 

[voko]

Again, the initial velocity is 80 km/h at 30 degree. You cannot count i t twice, it makes no sense.

 

[negation]

Again, the initial velocity is 80 km/h at 30 degree. You cannot count i t twice, it makes no sense.

I did contemplate about this quandary-and I'm sure you are right in your statement-but I was thinking along the line of 80kmh^-1 + velocity of 80kmh^-1 at 30 degree at the instantaneous moment when it skid to give vi...

 

[negation]

otherwise,

(-19.22ms^-1 , -11.1ms^-1)/3.9s = (-4.9ms^-2 , -2.846ms^-2)

 

[voko]

There is only one velocity at any given moment of time. If it is given to be 80 km/h at some angle, then its x component is 80 km/h times the cosine, and the y component is 80 km/h times the sine. You cannot say, oh, it was 80 km/h "straight" just a moment ago, let's add that to what it is now - because then you can just say, wait, it was some other value just one and a half moment ago, so let's add that as well - where do you stop in that sort of reasoning? One velocity at anyone time, plain and simple.

 

[voko]

otherwise,

(-19.22ms^-1 , -11.1ms^-1)/3.9s = (-4.9ms^-2 , -2.846ms^-2)

What magnitude of acceleration does that yield? Is that consistent with the other method?

 

[negation]

What magnitude of acceleration does that yield? Is that consistent with the other method?

Yes, this returns me the value 5.7ms^-2-consistent with the average acceleration.

 

[negation]

There is only one velocity at any given moment of time. If it is given to be 80 km/h at some angle, then its x component is 80 km/h times the cosine, and the y component is 80 km/h times the sine. You cannot say, oh, it was 80 km/h "straight" just a moment ago, let's add that to what it is now - because then you can just say, wait, it was some other value just one and a half moment ago, so let's add that as well - where do you stop in that sort of reasoning? One velocity at anyone time, plain and simple.

I anticipated this and was aware of the mathematical and logical implication..