A dog running in an open field has components of velocity v x = 2.9 m / s and v
ID: 1325156 • Letter: A
Question
A dog running in an open field has components of velocity vx = 2.9m/s and vy = -1.6m/s at time t1 = 12.0s . For the time interval from t1 = 12.0s to t2 = 23.5s , the average acceleration of the dog has magnitude 0.52m/s2 and direction 34.0? measured from the +x?axis toward the +y?axis.At time t2 = 23.5s , what is the x-component of the dog's velocity?
At time t2 = 23.5s , what is the y-component of the dog's velocity?
What is the magnitude of the dog's velocity?
What is the direction of the dog's velocity (measured from the +x?axis toward the +y?axis)?
A dog running in an open field has components of velocity vx = 2.9m/s and vy = -1.6m/s at time t1 = 12.0s . For the time interval from t1 = 12.0s to t2 = 23.5s , the average acceleration of the dog has magnitude 0.52m/s2 and direction 34.0? measured from the +x?axis toward the +y?axis.
At time t2 = 23.5s , what is the x-component of the dog's velocity?
At time t2 = 23.5s , what is the y-component of the dog's velocity?
What is the magnitude of the dog's velocity?
What is the direction of the dog's velocity (measured from the +x?axis toward the +y?axis)?
At time t2 = 23.5s , what is the x-component of the dog's velocity?
At time t2 = 23.5s , what is the y-component of the dog's velocity?
What is the magnitude of the dog's velocity?
What is the direction of the dog's velocity (measured from the +x?axis toward the +y?axis)? What is the direction of the dog's velocity (measured from the +x?axis toward the +y?axis)?
Explanation / Answer
Since units are not mentioned therefore I am assuming the SI units.
Average acceleration is given as 0.52 m/s2 and the direction is given as 34 therefore
Acceleration in X direction ax= 0.52Cos34 = 0.43 m/s2
Acceleration in y direction ay= 0.52 Sin34 = 0.29 m/s2
Now applying the first equation of motion we will calculate the final velocity in x and y direction .
In X direction Vfx = (Vix=2.9) + axt where t is the time interval (23-12 = 11 s) and Vfx is final velocity in x direction.
Vfx = 7.63 m/s
Similarly in y direction Vfy = 1.59 m/s
NOw for final velocity magmnitude apply pythagoras theorem = V2= Vfx2+Vfy2
V = 7.79 m/s
For direction tan(angle) = Vfy/Vfx
angle = 11.77
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