I'm stuck and need assistance(velocity/acceleration vector problem)

[tomdadude]

Homework Statement

A car initially traveling north at 35 m/s, brakes and turns to avoid an obstacle so
that 3 seconds later it is traveling northeast at 20 m/s. Find its average accelera-
tion during the period of braking and turning.

Homework Equations

acceleration=change in velocity/time taken(thank you Chi Meson)

The Attempt at a Solution

I got the vector components of the 20m/s vector and got Vx as 10m/s and Vy as 10 root 3 m/s.
I don't thinks it's as simple as 15m/s divided by 3s.Any help would be very appreciated.I don't know where to go from here.

I'm new here and I'd like to say hello to every member.

 

[Chi Meson]

The definition of acceleration is the "rate of change in velocity." (or "change in velocity" / time). Not the "average velocity/time."

Try this graphically: draw vector arrows for these velocities with their bases together. The change in velocity is the ve3ctor arrow that goes from the tip of the initial velocity to the tip of the final velocity. You can use simple trig to calculate it.

Furthermore: how did you get different x and y components for a vector pointing at 45 degrees?

 

[tomdadude]

It has changed by 15m/s.Sorry about that I'm getting all mixed up.I heard that I have to get the vector components of the acceleration so I need help there really.

 

[Chi Meson]

It has changed by 15m/s.Sorry about that I'm getting all mixed up.I heard that I have to get the vector components of the acceleration so I need help there really.

No, it has not changed by 15 m/s, even though it looks like it from a strictly 1-D mathematical viewpoint. The magnitude of the change in velocity is (believe it or not) more than 20 m/s.

Did you draw the arrows I suggested?

Or strictly with algebra: what is the change in velocity in the x-direction? \Delta v_x
what is the change in velocity in the y direction? \Delta v_y