finding angle of average velocity

Question

A football player runs for a distance d₁ = 8.57 m in 1.21 s, at an angle of θ = 62.9 degrees to the 50-yard line, then turns left and runs a distance d₂ = 10.58 m in 1.59 s, in a direction perpendicular to the 50-yard line. The diagram shows these two displacements relative to an x-y coordinate system, where the x axis is parallel to the 50-yard line, and the y axis is perpendicular to the 50-yard line.

What is the magnitude of the average velocity, in m/s?

What angle, in degrees does the average velocity make with the y axis? (Note that the angle θ was given as measured from the x axis rather than the y axis.)

Answer 1

Jule,

I applied trigonometry, as follows, to solve the problem. (BTW, I could not find the problem statement's referenced diagram.)

Step 1: Decompose the player's 62.9^° run into its x and y components:

H = Hypotenuse Distance = 8.57m

θ = 62.9^°

Vertical Distance Component = H * Sin(θ) = 7.63m (call this y1 - see Step 2, below)

Horizontal Distance Component = H * Cos(θ) = 3.904m (see Step 3, below)

Step 2: Determine total vertical distance of the run:

Total Vertical Distance = y1 + 10.58m = 7.63 + 10.58 = 18.21m (see Step 3, below)

Step 3: Determine Total Direct Distance and Total Time for the run (if it was made directly from Start Point to End Point):

Applying the Pythagorean Theorem:

Total Direct Distance = √(Horizontal Distance Component)² + (Total Vertical Distance)²

= √(3.904² + 18.21²) = 18.624m (ANS)

Total Time = 1.21s + 1.59s = 2.8s

Step 4: Determine the Average Velocity and Angle of the run (if it was made directly from Start Point to End Point):

Run Average Velocity = Total Direct Distance / Total Time = 18.624m / 2.8s = 6.65m/s (ANS)

Note: True Velocity = Sum of Distances / Total Time = 19.15m / 2.8s = 6.84m/s

θ_avgX = The angle of the run w/r/t the x-axis = ArcTan(18.21 / 3.904) = 77.9^°

θ_avgY = The angle of the run w/r/t the y-axis = 90^° - 77.9^° = 12.1^° (ANS)

I hope this helps you.

Gary