Acceleration and velocity (conceptual) question

[mateomy]

I was doing some problems out of my book and the following problem has been solved, but a little question popped up in my head. Its simply conceptual, just want to gain a deeper and truer understanding of these two fundamental ideas...

The problem goes something like this...(easy one)...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 3.9s. 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 answer that I stumbled upon AFTER having figured out the individual components of the resultant vector is; 5.69 m/s^2.My question is: Why is it that we only take into account the acceleration coming from the x coordinate? The car doesn't stay simply along the x-axis as is noticed from the 30(degree) "skid" it makes across the road.

Shouldn't the average acceleration be figured from the resultant vector? Or is the book just leaving out specifics?

 

[Delphi51]

The initial AND final y-component velocities are zero, so you could say the average is zero. There is no way to find a value for the acceleration unless given the maximum sideways speed.

 

[mateomy]

Sorry if this is simple-minded, but what then, causes the 30 degree skid?

 

[androidx219]

Ideally if the car is strictly driving straight to the x-axis and the brakes are applied on both the wheels with identical force, then the car shouldn't rotate at all (assuming the slippery road has no geometrical flaws). In reality this situation might be hard to achieve. So any unbalanced force arising with the difference of brakes on both wheels or slightly wrong alignment might produce a nonzero torque on the car that will start rotating it. After that its all at hands of inertia and Newtonian mechanics.

 

[mateomy]

That makes it much more understandable. Thank you both.