/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Let the distance travelled by ca... [FREE SOLUTION] | 91影视

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Let the distance travelled by car with \(42 \mathrm{~km} \mathrm{~h}^{-1}\) for 20 minutes be \(\mathrm{S}\), and distance travelled with 60 \(\mathrm{km} \mathrm{h}^{-1}\) for 30 minutes be \(\mathrm{S}_{2}\) \(\Rightarrow \mathrm{S}_{1}=42 \mathrm{~km} \mathrm{~h}^{-1} \times \frac{1}{3} \mathrm{~h}=14 \mathrm{~km}\) \(\Rightarrow \mathrm{S}_{2}=60 \mathrm{~km} \mathrm{~h}^{-1} \times \frac{1}{2} \mathrm{~h}=30 \mathrm{~km}\) Total distance \(\mathrm{S}=\mathrm{S}_{1}+\mathrm{S}_{2}=14+30=44 \mathrm{~km}\)

Short Answer

Expert verified
Answer: The total distance traveled by the car is 44 km.

Step by step solution

01

Calculate the first distance (S鈧)

To calculate the first distance (S鈧), we need to multiply the speed (42 km/h) with the time (20 minutes or 1/3 hours). So, we have: S鈧 = 42 km/h 脳 (1/3) h
02

Calculate the second distance (S鈧)

To calculate the second distance (S鈧), we need to multiply the speed (60 km/h) with the time (30 minutes or 1/2 hours). So, we have: S鈧 = 60 km/h 脳 (1/2) h
03

Find the total distance (S)

Now that we've calculated both distances S鈧 and S鈧, we can find the total distance (S) by adding them together: S = S鈧 + S鈧 Now, we will perform the calculations and find the total distance.
04

Calculate S鈧

Let's calculate the first distance (S鈧) based on the formula we've written in Step 1: S鈧 = 42 km/h 脳 (1/3) h = 14 km
05

Calculate S鈧

Let's calculate the second distance (S鈧) based on the formula we've written in Step 2: S鈧 = 60 km/h 脳 (1/2) h = 30 km
06

Calculate the total distance (S)

Now we will find the total distance traveled (S) according to the formula in Step 3. Adding the results from Step 4 and Step 5, we get: S = S鈧 + S鈧 = 14 km + 30 km = 44 km The total distance traveled by the car is 44 km.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed and Time
Understanding the relationship between speed and time is essential when calculating distance. Speed tells us how much distance is covered over a certain period of time. If you're driving at a speed of 42 kilometers per hour, for example, it means you're traveling 42 kilometers in one hour. This concept helps us to determine how far you can go in a different amount of time.
To find the distance, we multiply the speed by the time. For instance, if we need to find out how far a car travels in 20 minutes at a speed of 42 kilometers per hour, we first convert the 20 minutes into hours because speed is given in kilometers per hour.
  • 20 minutes = \(\frac{20}{60}\) hours = \(\frac{1}{3}\) hours
In this scenario, the distance covered in 20 minutes can be calculated using the formula:
  • Distance = Speed \(\times\) Time = 42 km/h \(\times\) \(\frac{1}{3}\) h = 14 km
Understanding this relationship enables us to solve for distance whenever we know the speed and time, demonstrating the practical application of this concept in everyday situations.
Distance Formula
The distance formula is a simple yet powerful tool to calculate how far an object travels. The formula is:
  • Distance = Speed \(\times\) Time
This formula is derived from the basic definition of speed, where speed is the rate at which distance is covered. The formula is highly versatile and can be used in various scenarios, whether you are walking, biking, or driving.
In our exercise example, we used this formula to calculate two separate distances. First, for a car traveling at 42 km/h for 20 minutes, we found
  • Distance \(S_1\) = 42 km/h \(\times\) \(\frac{1}{3}\) h = 14 km
Then, for another journey at 60 km/h for 30 minutes, we calculated
  • Distance \(S_2\) = 60 km/h \(\times\) \(\frac{1}{2}\) h = 30 km
Finally, the overall distance covered (S = \(S_1 + S_2\)) was found to be 44 kilometers. By applying the distance formula, you can swiftly and accurately determine how far you've traveled or plan to travel.
Units Conversion
Converting units is crucial when working with formulas involving speed, distance, and time. Often, the time is given in minutes while speed is given in hours, requiring a conversion to keep units consistent.
Let's consider our exercise where we needed to convert minutes into hours:
  • 20 minutes is converted to hours by dividing by 60 (since there are 60 minutes in an hour): \(\frac{20}{60} = \frac{1}{3}\) hours
  • Similarly, 30 minutes is converted as \(\frac{30}{60} = \frac{1}{2}\) hours
By standardizing time to hours, calculations become straightforward, and we can apply the distance formula without discrepancies. Each type of measurement has its specific unit, and conversions like these are foundational in ensuring calculations are correct.
Remember, always check your units to ensure that they are compatible when applying mathematical formulas; this habit will help avoid errors in calculations.

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Most popular questions from this chapter

The initial velocity, \(\mathrm{u}=108 \mathrm{~km} \mathrm{~h}^{-1}=30 \mathrm{~m} \mathrm{~s}^{-1}\) The final velocity, \(\mathrm{v}=0 \mathrm{~m} \mathrm{~s}^{-1}\) The time taken to stop lorry, \(\mathrm{t}=30 \mathrm{~s}\). The acceleration of the lorry, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{0-30}{30}=-1 \mathrm{~m} \mathrm{~s}^{-2}\)

Unit of time in SI and CGS systems is second (s). Unit of speed in SI and CGS systems are \(\mathrm{m} \mathrm{s}^{-1}\) and \(\mathrm{cm} \mathrm{s}^{-1}\) respectively.

(i) The unit of speed and velocity is \(\mathrm{m} \mathrm{s}^{-1}\). (ii) If a body is moving in a straight line path then its speed is equal to velocity \(50 \mathrm{~m} \mathrm{~s}^{-1}=50 \times \frac{18}{5} \mathrm{~km} \mathrm{~h}^{-1}=180 \mathrm{~km} \mathrm{~h}^{-1}\)

The number of oscillations completed in one second is called its frequency. The bob is oscillating once in every 6 seconds The number of oscillations in \(1 \mathrm{~s}=1 / 6\)

\(\mathrm{A} \rightarrow \mathrm{b} \quad\) The piston of a motorcar engine moving at uniform speed is said to be in periodic motion. \(\mathrm{B} \rightarrow \mathrm{e}, \mathrm{b} \quad\) The objects executing vibratory motion undergo change in shape or size. The piston of a motor car engine executes vibratory motion. \(\mathrm{C} \rightarrow \mathrm{g} \quad\) Body at rest will have zero speed as well as zero velocity. \(\mathrm{D} \rightarrow\) a Maximum displacement of a body from its mean position is called amplitude. \(\mathrm{E} \rightarrow \mathrm{c} \quad\) A body moving with variable speed is said to be in non-uniform motion. \(\mathrm{F} \rightarrow \mathrm{d} \quad 1 \mathrm{~ms}^{-1}=\frac{1 \mathrm{~m}}{1 \mathrm{~s}}=\frac{\frac{1}{100} \mathrm{~km}}{\frac{1}{3600} \mathrm{~h}}=\frac{18}{5} \mathrm{~km} \mathrm{~h}^{-1}\) \(\mathrm{G} \rightarrow \mathrm{f} \quad\) Average velocity \(=\frac{\text { Total displacement }}{\text { Total time }}\)

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