Chapter 2, Problem 81GP

To determine

The speed of conveyor belt

Answer to Problem 81GP

  $0.44\frac{\text{m}}{\text{min}}$

  $3\frac{\text{burgers}}{\min }$

Explanation of Solution

Given:

Total distance travelled by burger in the grilling machine = length of the grilling machine $=d_c=1.1\text{m}$

Time of travel $=t=2.5\min$

Speed of conveyor belt $=v_c$

Length of grilling machine travelled $=d$

Distance between two burgers $=D=15\text{cm}=0.15\text{m}$

Time $=t=1\min$

Formula Used:

Speed is given as

  $\text{Speed = }\frac{\text{distance}}{\text{time}}$

Distance travelled is given as

  $\text{distance = (speed)(time)}$

Rate or production is given as

  $\text{Rate = }\frac{\text{distance traveled on the belt}}{\text{distance between two burgers}}$

Calculation:

Speed of conveyor belt is given as

  $\begin{array}{l}{v}_c=\frac{d_c}{t}\\ v_c=\frac{1.1}{2.5}\\ v_c=0.44\frac{\text{m}}{\text{min}}\end{array}$

Distance travelled on the conveyor belt is given as

  $\begin{array}{l}d=v_{c}t\\ d=(0.44)\times 1\\ d=0.44\text{m}\end{array}$

Rate of production of burger is given as,

  $R=\frac{d}{D}$

Substituting the values.

  $\begin{array}{c}\text{R = }\frac{0.44}{0.15}\\ R=2.93\frac{\text{burgers}}{\min }\\ \approx 3\frac{\text{burgers}}{\min }\end{array}$

Conclusion:

Therefore, speed of the conveyor belt is $0.44\frac{\text{m}}{\text{min}}$

The rate of burger production is $3\frac{\text{burgers}}{\min }$