Answer:
2.375 m, 54°
Explanation:
I will be doing this applying the vector format of i and j. I hope that's not confusing for you. Where, i = East and j = North.
First, we convert the velocity from miles per hour to meters per second. Remember that 1 mile = 1609 meters
u = 1 mph East = 1.00i + 0j mph
u = (1.00i + 0j) mph * (1609 m / 1 mile) * ( 1 hr / 3600 s) = 0.45i + 0j m/s
a = 0.4 m/s² FROM THE SOUTH
a = 0.4 m/s² North = 0i + 0.4j m/s²
Next, we proceed to use the equation of motion to solve
sf = si + ut + 1/2at²
Let's assume that si = 0, this means that
sf = ut + 1/2at², now we substitute for u, t and a to get
sf = (0.45i + 0j m/s) * 3.1 s + 1/2 * (0i + 0.4j m/s²) * (3.1s)²
sf = (1.395i + 0j m) + (0+1.922j m)
sf = 1.395i + 1.922j m
This means that after 3.1 s, the birds displacement from the start is 1.395 m in the East direction and 1.922 m in the North direction.
Finding the resultant now will be
|sf| = √(1.395² + 1.922²) m
|sf| = √(1.946 + 3.694) m
|sf| = √5.64
|sf| = 2.375 m
direction = Tanθ = (j/i)
θ = Tan^-1 (j/i)
θ = Tan^-1(1.922/1.395)
θ = Tan^-1(1.378)
θ = 54° North East
