
Out of a total distance of $200\,km$ covered, the first half is covered at the speed of $60\,km/hr$ and the remaining half is covered at $75\,km/hr$. Find the average speed for the whole journey.
A. $21\dfrac{2}{3}$ km/hr
B. $66\dfrac{2}{3}$ km/hr
C. $43\dfrac{2}{3}$ km/hr
D. $39\dfrac{2}{3}$ km/hr
Answer
498.6k+ views
Hint:The pace at which an object's location changes in any direction is known as speed. Speed is defined as the ratio of distance travelled to the time it took to travel that distance. Because speed has just one direction and no magnitude, it is a scalar number. The following notion is employed in this instance.
Complete step by step answer:
The magnitude of the rate of change of an item's position with time or the magnitude of change of its position per unit of time is the speed (often referred to as v) of an object in daily use and in kinematics; it is therefore a scalar number. The average speed of an object in a given time interval is the item's distance travelled divided by the period's duration; the instantaneous speed is the average speed's limit as the interval's duration approaches zero.
$\text{Speed} = \dfrac{{{\text{ Distance }}}}{{{\text{ Time }}}}$
$\Rightarrow \text{AverageSpeed} = \dfrac{{{\text{ Total Distance Travelled }}}}{{{\text{ Total Time Taken }}}}$
Since 100 Km is covered at 60 km/hr.Time,
${{\text{t}}_1} = \dfrac{{{\text{ Distance }}}}{{{\text{ Speed }}}} \\
\Rightarrow {{\text{t}}_1} = \dfrac{{100}}{{60}} \\
\Rightarrow {{\text{t}}_1} = \dfrac{5}{3}h$
And since another 100 km is covered at 75 km/hr. Time,
${{\text{t}}_2} = \dfrac{{{\text{ Distance }}}}{{{\text{ Speed }}}} \\
\Rightarrow {{\text{t}}_2} = \dfrac{{100}}{{75}} \\
\Rightarrow {{\text{t}}_2} = \dfrac{4}{3}$
So, the total time taken is represented as
$t = \dfrac{5}{3} + \dfrac{4}{3} = 3{\text{hr}}$
Hence, average speed
$\text{AverageSpeed} = \dfrac{{200}}{3} \\
\therefore \text{AverageSpeed}= 66\dfrac{2}{3}\;{\text{km h}}{{\text{r}}^{ - 1}}$
Hence, the correct answer is option B.
Note:Make sure you follow the correct unit conversions. The parameters of speed are distance divided by time. The metre per second (m/s) is the SI unit of speed, although the kilometre per hour (km/h) or miles per hour (in the US and the UK) is the most frequent measure of speed in everyday use (mph). The knot is frequently used in aviation and maritime transport.
Complete step by step answer:
The magnitude of the rate of change of an item's position with time or the magnitude of change of its position per unit of time is the speed (often referred to as v) of an object in daily use and in kinematics; it is therefore a scalar number. The average speed of an object in a given time interval is the item's distance travelled divided by the period's duration; the instantaneous speed is the average speed's limit as the interval's duration approaches zero.
$\text{Speed} = \dfrac{{{\text{ Distance }}}}{{{\text{ Time }}}}$
$\Rightarrow \text{AverageSpeed} = \dfrac{{{\text{ Total Distance Travelled }}}}{{{\text{ Total Time Taken }}}}$
Since 100 Km is covered at 60 km/hr.Time,
${{\text{t}}_1} = \dfrac{{{\text{ Distance }}}}{{{\text{ Speed }}}} \\
\Rightarrow {{\text{t}}_1} = \dfrac{{100}}{{60}} \\
\Rightarrow {{\text{t}}_1} = \dfrac{5}{3}h$
And since another 100 km is covered at 75 km/hr. Time,
${{\text{t}}_2} = \dfrac{{{\text{ Distance }}}}{{{\text{ Speed }}}} \\
\Rightarrow {{\text{t}}_2} = \dfrac{{100}}{{75}} \\
\Rightarrow {{\text{t}}_2} = \dfrac{4}{3}$
So, the total time taken is represented as
$t = \dfrac{5}{3} + \dfrac{4}{3} = 3{\text{hr}}$
Hence, average speed
$\text{AverageSpeed} = \dfrac{{200}}{3} \\
\therefore \text{AverageSpeed}= 66\dfrac{2}{3}\;{\text{km h}}{{\text{r}}^{ - 1}}$
Hence, the correct answer is option B.
Note:Make sure you follow the correct unit conversions. The parameters of speed are distance divided by time. The metre per second (m/s) is the SI unit of speed, although the kilometre per hour (km/h) or miles per hour (in the US and the UK) is the most frequent measure of speed in everyday use (mph). The knot is frequently used in aviation and maritime transport.
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