The average distance per unit time, when the body is moving with variable speed, is called average speed, Average speed \(=\frac{\text { Total distance travelled }}{\text { Total time taken }}\)

Short Answer

Expert verified
Answer: To find the average speed, follow the steps outlined in the solution: Step 1: Identify the total distance traveled and total time taken. The car travels 40 km in the first hour and 60 km in the second hour, so the total distance traveled is 40 km + 60 km = 100 km. The total time taken is 1 hour + 1 hour = 2 hours. Step 2: Substitute values into the average speed formula \(\text{Average speed} = \frac{\text{ Total distance travelled }}{\text{ Total time taken }} = \frac{100 \,\text{km}}{2\,\text{hours}}\) Step 3: Solve for average speed Divide the total distance (100 km) by the total time (2 hours) to find the average speed: \(\text{Average speed} = 50\,\text{km/h}\) Step 4: Determine the units The average speed will be in kilometers per hour (km/h) since distance is in kilometers and time is in hours. Therefore, the average speed is 50 km/h. Step 5: Interpret your result The car's average speed was 50 km/h over the two-hour period. This means that, on average, the car was moving at a speed of 50 km/h throughout its journey.

Step by step solution

01

Identify the total distance traveled and total time taken

To use the formula for average speed, you need to first find the total distance traveled and the total time taken. This can be done by analyzing the problem statement, looking at graphs, or using equations for distance and time. Make sure both distance and time are in the same units.
02

Substitute values into the average speed formula

Once the total distance traveled and total time taken have been identified, plug these values into the formula: \(\text{Average speed} = \frac{\text { Total distance travelled }}{\text { Total time taken }}\). Replace "Total distance traveled" with the actual distance and "Total time taken" with the actual time, keeping the units consistent.
03

Solve for average speed

When both values, distance and time, are plugged into the formula, perform the division to find the average speed. Divide total distance by total time to obtain the numerical value for average speed.
04

Determine the units

The units for average speed will be the same as the units used for distance and time. Commonly used units for average speed are kilometers per hour (km/h), meters per second (m/s), or miles per hour (mph). Attach the appropriate unit to your numerical value to express the average speed.
05

Interpret your result

Once you have calculated the average speed, make sure to give a brief interpretation of your result. For example, you can explain what the average speed represents in the context of the given problem. How fast was the body moving on average over the given time?

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Most popular questions from this chapter

Fill in the Blanks. \(\frac{\mathrm{T}}{2}\) The time taken to perform one to-and-fro motion (or) from one extreme position to other extreme position and back is called time period (T).

(a) Let the distance travelled be 'd'. The speed of \(^{\prime} A^{\prime}, v_{A}=\frac{d}{20} m s^{-1}\) The speed of ' \(\mathrm{B}^{\prime}, \mathrm{v}_{\mathrm{B}}=\frac{\mathrm{d}}{22} \mathrm{~m} \mathrm{~s}^{-1}\). \(\Rightarrow \frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\frac{\mathrm{d}}{20}}{\frac{\mathrm{d}}{22}}=\frac{22}{20}=\frac{11}{10}=11: 10\) (b) Let them run for 't's. Then, \(\mathrm{d}_{\mathrm{A}}=\mathrm{v}_{\mathrm{A}} \times \mathrm{t}=\frac{\mathrm{d}}{20} \times \mathrm{t}\) \(\mathrm{d}_{\mathrm{B}}=\mathrm{v}_{\mathrm{B}} \times \mathrm{t}=\frac{\mathrm{d}}{22} \times \mathrm{t}\) \(\Rightarrow \frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=\frac{\frac{\mathrm{d}}{20}(\mathrm{t})}{\frac{\mathrm{d}}{22}(\mathrm{t})}=\frac{22}{20}=\frac{11}{10}=11: 10\)

Fill in the Blanks. vibratory The molecules in solid undergo vibratory motion.

Fill in the Blanks. \(\frac{1}{1000}\) 1000 millisecond \(=1 \mathrm{~s}\) \(\Rightarrow 1\) millisecond \(=\frac{1}{1000}\) th part of a second.

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, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{0-30}{30}=-1 \mathrm{~m} \mathrm{~s}^{-2}\)

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