Question

An unsuspecting bird is coasting along in an easterly direction at 1.00 mph1.00 mph when a strong wind from the south imparts a constant acceleration of 0.400 m/s20.400 m/s2 . If the wind's acceleration lasts for 3.10 s3.10 s , find the magnitude rr and direction θθ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally travelling in the +x+x direction and that there are 1609 m1609 m in 1 mi1 mi .

Answer 1

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