Chapter 3, Problem 5CLR

To determine

To calculate: The time spent while driving at 55mi/h.

Answer to Problem 5CLR

The time spent while driving at 55mi/h is 4 hours.

Explanation of Solution

Given information: On a 5-hour trip, some part is travelled at 30mi/h and rest at 55mi/h. Total distance travelled is 250mi.

Calculation:

Formula used: $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Denote the distance covered at 30mi/h by x and the distance covered at 55mi/h by y.

Time taken to drive at the speed of 30mi/h is,

  $\begin{array}{c}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ 30=\frac{x}{{\text{T}}_1}\\ {\text{T}}_1=\frac{x}{30}\end{array}$

• Here, $T_1$ represents the time taken to drive at the speed of 30mi/h.

Time taken to drive at the speed of 55mi/h is,

  $\begin{array}{c}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ 55=\frac{y}{{\text{T}}_2}\\ {\text{T}}_2=\frac{y}{55}\end{array}$

• Here, $T_2$ represents the time taken to drive at the speed of 55mi/h.

Total time spent on the journey is 5-hours. Therefore,

  $\begin{array}{c}{\text{T}}_1+{\text{T}}_2=5\\ \frac{x}{30}+\frac{y}{55}=5\left({\text{Substitute the values of T}}_1\text{ and }{T}_2\right)\\ \frac{x}{5\times 6}+\frac{y}{5\times 11}=5\\ \frac{11x+6y}{5\times 6\times 11}=5\left(\text{L}\text{.C}\text{.M}\text{. of 30 and 55 is 330}\right)\\ 11x+6y=1650\mathrm{......}\left(1\right)\end{array}$

Total distance covered is given to be 250mi. Therefore,

  $\begin{array}{c}\text{Total distance}=250\\ x+y=250\\ y=250-x\mathrm{......}\left(2\right)\end{array}$

Substitute the value of y from equation(2) in equation (1) and simplify,

  $\begin{array}{c}11x+6y=1650\\ 11x+6\left(250-x\right)=1650\left(\because y=250-x\right)\\ 11x+1500-6x=1650\\ 5x=1650-1500\\ 5x=150\\ x=\frac{150}{5}\\ =30\end{array}$

Since y represents the distance covered while driving at 55mi/h, the distance covered is,

  $\begin{array}{c}y=250-x\\ =250-30\\ =220\end{array}$

Time spent while driving at 55mi/h speed is,

  $\begin{array}{c}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ 55=\frac{y}{{\text{T}}_2}\\ {\text{T}}_2=\frac{y}{55}\\ =\frac{220}{55}\\ =4\end{array}$

Hence, 4 hours are spent while travelling at the speed of 55mi/h.