First we will examine the motion of the suitcase relative to the truck:
For acceleration:
ax - ax
o = (1/2)v
2 - (1/2)v
o2
a = (v
2 - v
o2) / 2(x - x
o) = 15
2 - 0 / (2*(10-0))
= 11.25 ft/s
2
For time:
at = v - v
0
t = (v - v
o) / a
= 15 - 0 / 11.25 = 1.3333 seconds
So the velocity at t = 1.2 seconds
v = v0 + at
= 0 + (11.25)(1.2) = 13.5 ft/s
In vector form:
vs/t = 13.5 ft/s > 20o (> is the reference angle, not an inequality)
Now for the motion of the truck:
convert 6in = .5 ft
For the acceleration:
x = xo + vot + (1/2)at2
.5 = 0 + 0 + (1/2)a(1.33332)
a = (2*.5) / (1.3332) = .5625 ft/s2
The truck's velocity at t=1.2 seconds will be:
v = vo + at
= 0 + .5625*1.2 = .675 ft/s
In vector form: vt = .675 ft/s --->
The velocity of the suitcase at t=1.2 seconds will require some trig:
vs = vt + vs/t
Using the law of cosines:
vs2 = (13.52) + (.675)2 - (2)(13.5)cos20
= 165.579 = 12.87 ft/s
Now, using the law of sines:
.675/ sinα = vs / sin20
sinα = .675(sin20) / 12.87 = .01794
α = sin-1(.01794)
α = 1.03o β = 20 + 1.03 = 21.03o
so vs = 12.87 ft/s > 21.03o
Note: The > and < represent reference angles, not inequalities