2-D Kinematic Velocity and Acceleration Practice Problem

[Panda1321]

"A dog running in an open field has components of velocity vx = 1.8 m/s and vy = -1.8 m/s at time t1 = 10.8 s . For the time interval from t1 = 10.8 s to t2 = 23.2 s , the average acceleration of the dog has magnitude 0.45 m/s2 and direction 32.0 ∘ measured from the +x−axis toward the +y−axis."

A. "At time t2 = 23.2 s , what is the x-component of the dog's velocity?"

Equations:
Vx= dx/dt Vy= dy/dt First, I tried to find the velocity at t2=23.2s by: V=at V=(0.45m/s^2)(23.2s)=10.44m/s
Therefore Vx=10.44m/s

However, this answer is incorrect and I am unable to understand how to find Vx with the data given. Can someone please explain how to solve this problem and if the magnitude and direction given from T1 to T2 is different than the angle formed by starting at the origin to T1? Thanks!

 

[Orodruin]

You have assumed that the acceleration is completely in the x direction. According to the problem statement, this is not true.

Edit: Furthermore, you are not accounting for the initial velocity or the fact that the acceleration did not start at t=0.

 

[NFuller]

First, I tried to find the velocity at t2=23.2s by: V=at V=(0.45m/s^2)(23.2s)=10.44m/s
Therefore Vx=10.44m/s

However, this answer is incorrect and I am unable to understand how to find Vx with the data given. Can someone please explain how to solve this problem and if the magnitude and direction given from T1 to T2 is different than the angle formed by starting at the origin to T1? Thanks!

There are several problems. As @Orodruin said you need to work with both the x and y components of velocity.

You also forgot about the initial velocity. Remember the 1-d motion equation is $v=v_{0}+at$.

Finally the acceleration is not for the entire 23.2s as you have used it. The dog only accelerates for some of that time.