Q87P - Fundamentals Of Physics

Question

When the legal speed limit for New York thruway was increased from $\mathbf{55}\mathbf{ }\mathbf{mi}\mathbf{/}\mathbf{h}$ to$\mathbf{65}\mathbf{ }\mathbf{mi}\mathbf{/}\mathbf{h}$ , how much time was saved by a motorist who drove the 700 km between the buffalo entrance and the New York city exit at the legal speed limit?

Step-by-Step Solution

Answer

The time saved by the motorist who drove the700 km between the Buffalo entrance and the New York City exit at the legal speed limit is 1h 13 min

Step 1: Given information

Distance between the buffalo entrance and the New York city  $\mathrm{x}=700 \mathrm{km}$

$$\begin{gathered} {\mathrm{v}}_1=55\frac{\mathrm{mi}}{\mathrm{h}} \\ {\mathrm{v}}_2=65\frac{\mathrm{mi}}{\mathrm{h}} \end{gathered}$$

Step 2: To understand the concept. Relation between speed, distance and time at constant acceleration

This problem involves simple physical quantities speed, distance, and acceleration. When acceleration is constant one can say that speed is the ratio of distance and time.

Formula:

The velocity is given by,

$\mathrm{\Delta v}=\frac{\mathrm{x}}{\mathrm{\Delta t}}$

Step 3: Calculations for time saved by a motorist

$$\begin{gathered} {\mathrm{v}}_1=55\frac{\mathrm{mi}}{\mathrm{hr}} \\ =55\frac{\mathrm{mi}}{\mathrm{hr}}\times \left(\frac{1609 \mathrm{m}}{1 \mathrm{mi}}\right)\times \left(\frac{1 \mathrm{hr}}{3600 \sec }\right) \\ =24.58 \mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} {\mathrm{v}}_2=65\frac{\mathrm{mi}}{h} \\ =65\frac{\mathrm{mi}}{\mathrm{hr}}\times \left(\frac{1609 \mathrm{m}}{1 \mathrm{mi}}\right)\times \left(\frac{1 \mathrm{hr}}{3600 \sec }\right) \\ =29.05 \mathrm{m}/\mathrm{s} \end{gathered}$$

By the definition of velocity

$$\begin{gathered} \mathrm{\Delta v}=\frac{\mathrm{x}}{\mathrm{\Delta t}} \\ \mathrm{\Delta t}=\frac{\mathrm{x}}{\mathrm{\Delta v}} \end{gathered}$$

The time saved by the motorist who drove the 700 km between the Buffalo entrance and the New York City exit at the legal speed limit is 

$$\begin{gathered} \mathrm{\Delta t}={\mathrm{t}}_1-{\mathrm{t}}_2 \\ \mathrm{\Delta t}=\frac{\mathrm{x}}{{\mathrm{v}}_1}-\frac{\mathrm{x}}{{\mathrm{v}}_2} \\ \mathrm{\Delta t}=\frac{700\times {10}^3\mathrm{m}}{24.58 \mathrm{m}/\mathrm{s}}-\frac{700\times {10}^3\mathrm{m}}{29.05 \mathrm{m}/\mathrm{s}} \\ \mathrm{\Delta t}=4383 \mathrm{s} \\ =73 \min \\ =1 \mathrm{h}13 \min \\ =1.2 \mathrm{hr} \end{gathered}$$

So, the time saved by the motorist is $1.2 \mathrm{hr}$