- #1
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!
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!