Answer:
a = 7.35 ft / s²
Explanation:
For this exercise we must use the kinematics relations
x = v₀ t + ½ a t²
as the runner leaves the starting line his initial velocity is zero
x = ½ a t²
a =
let's reduce the distance to foot
x = 60 yd (3ft / 1yd) = 180 ft
let's calculate
a = 2 180 / 7²
a = 7.35 ft / s²
consider east-west direction along x-axis and north-south direction along y-axis
from the diagram
A = Ax i + By j = - (0.40 Sin60) i + (0.40 Cos60) j = - 0.35 i + 0.20 j
B = Bx i + By j = - 0.50 i + 0 j
Net displacement is given as
D = A + B
D = (- 0.35 i + 0.20 j ) + (- 0.50 i + 0 j )
D = - 0.85 i + 0.2 j
magnitude of displacement is given as
|D| = sqrt((- 0.85)² + (0.2)²)
|D| = 0.87 km
direction of displacement is given as
θ = tan⁻¹(0.2/0.85)
θ = 13.5 deg north of west
