Question

. Boxes are sitting on a conveyor belt as the conveyor is turned on, moving the boxes toward the right. The belt reaches full speed of 45 fpm (ft/min) in 0.5 s. Determine the linear acceleration of the boxes assuming that this acceleration is constant. Also determine the linear displacement of the boxes during this speed-up period.

Answer 1

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.