The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.
The distance of car from the intersection point after t hours is .
The distance of truck from the intersection point after t hours is .
Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is
Thus the distance apart is
consider the motion along the X-direction
X = horizontal displacement = 80 m
= initial velocity along the x-direction = v Cos60
t = time of travel
using the equation
X = t
80 = (v Cos60) (t)
t = 160/v eq-1
consider the motion in vertical direction :
Y = vertical displacement = 20 m
= initial velocity in Y-direction = v Sin60
a = acceleration = - 9.8 m/s²
t = time of travel = 160/v
using the equation
Y = t + (0.5) a t²
20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²
v = 32.5 m/s