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Response Details:
v = 15 ft/s, s = 10 ft, d = 6" = 0.5 ft, θ = 20o, t' = 1.2 s
The motion of the suitcase relative to the truck:
initial velocity = 0, final velocity = v, acceleration = a, displacement = s, time = t
a = v2/(2s), direction: along the inclined plane
t = 2s/v
The motion of the truck:
initial velocity = 0, acceleration = a', displacement = d, time = t = 2s/v
d = a't2/2
so a' = 2d/t2 = 2d/(2s/v)2 = v2d/(2s2) = a*(d/s) direction: to right
at time t' = 1.2 s,
velocity of the suitcase relative to the truck is v = at' = v2t'/(2s) direction: along the inclined plane
velocity of the truck is v' = a't' = v2t'/(2s)*(d/s) direction: to right
velocity of the suitcase V = vector sum of v and v'
in horizontal direction: (left is positive) Vx = vcosθ - v' = v2t'/(2s) (cosθ - d/s)
in vertical direction: (down is positive) Vy = vsinθ = v2t'/(2s) sinsθ
magnitue of V = √(Vx2 + Vy2) = v2t'/(2s) √[(cosθ - d/s)2 + sin2θ]
plug all known values and get V = 12.87 ft/s2
angle with horizontal φ:
φ = arctan(Vy/Vx) = arctan[sinθ/(cosθ - d/s)] = 21.0o
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