Question

A conveyor belt is used to send burgers through a grilling machine. If the grilling machine is 1.2 m long and the burgers require 2.8 min to cook, how fast must the conveyor belt travel? If the burgers are spaced 25 cm apart, what is the rate of burger production (in burgers/min)?

Short answer

The obtained rate of burger production is $N=1.712\mathrm{burgers}/\min$.

Step-by-step solution

Step 1. Calculation of time

The average velocity can be calculated using the fraction between the total distance covered by the conveyor belt and the total time taken.

To calculate the rate of burger production, divide the average velocity with the separation distance among burgers.

Given data.

The length of the grilling machine is $l=1.2\mathrm{m}$.

The time required for cooking is $t=2.8\min$.

The distance between burgers is $d=25\mathrm{cm}$.

The relation used to find the average velocity is given by.

$v=\frac{l}{t}$

On plugging the values in the above relation, you get.

$\begin{array}{l}v=\left(\frac{1.2\mathrm{m}}{2.8\min }\right)\\ v=0.428\mathrm{m}/\min \end{array}$

Step 2. Calculation of the number of poles

The relation used to calculate the rate of burger production is given by.

$N=\left(\frac{1\mathrm{burger}}{d}\right)\left(v\right)$

On plugging the values in the above relation, you get.

$\begin{array}{l}N=\left(\frac{1\mathrm{burger}}{25\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}}\right)\left(0.428\mathrm{m}/\min \right)\\ N=1.712\mathrm{burgers}/\min \end{array}$

Thus, the rate of burger production is $N=1.712\mathrm{burgers}/\min$.