Answer:
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Answer:
The acceleration of the boxes is 1.5 ft/s²
The displacement of the boxes during the speed-up period is 0.1875 ft.
Explanation:
Hi there!
Let´s convert the 45 ft/min into ft/s:
45 ft/min · 1 min/ 60 s = 0.75 ft/s
It takes the belt 0.5 s to reach this speed. Then, the acceleration of the boxes will be:
a = v/t
Where:
a = acceleration.
v = velocity.
t = time.
a = 0.75 ft/s / 0.5 s
a = 1.5 ft/s²
The acceleration of the boxes is 1.5 ft/s²
The equation of displacement is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the boxes at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
Since the origin of the frame of reference is located at the point where the boxes begin to move, x0 = 0. Since the boxes were initially at rest, v0 = 0. Then:
x = 1/2 · a · t²
x = 1/2 · 1.5 ft/s² · (0.5 s)²
x = 0. 1875 ft
The displacement of the boxes during the speed-up period is 0.1875 ft.
m1= mass 1 = 1.1 kg
Vi1 = initial velocity 1 = 2.7 m/s
m2= 2.4 kg
V2i = -1.9 m/s
We assume east as positive and west as negative.
Apply the formulas:
Vf1 = ?
Replacing:
Answer: 3.6 m/s west
The answer to it is the letter A
