Question

Suppose you drive to Delhi \[\left( {200{\text{ }}Km{\text{ }}away} \right)\]at \[400\dfrac{{km}}{{hr}}\] and return\[200\dfrac{{km}}{{hr}}\] . What is your average speed for the entire trip?
(A) \[less{\text{ }}than300\dfrac{{km}}{{hr}}\]
(B) $More{\text{ }}than300\dfrac{{km}}{{hr}}$
(C) \[0\]
(D) $300\dfrac{{km}}{{hr}}$

Answer 1

Hint:To solve this question,we are going to apply the concept of average velocity and it is given as average speed \[ = \] total distance traveled \[/\]total time taken.
We can also directly calculated by using the formula;
\[v = \dfrac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\].

Complete step by step answer:
Average speed of an object is the total distance covered by the object divided by the elapsed time to cover the distance.
Average speed \[ = \] total distance traveled \[/\]total time taken.
Given,
Distance to Delhi \[ = {\text{ }}200{\text{ }}km.\]
Let us take \[{d_1}\]is the distance covered by the object while going and \[{d_2}\] is the distance covered by the object while returning.
Speed while going to Delhi \[ = 400\dfrac{{km}}{{hr}}\], Speed while returning is \[ = 200\dfrac{{km}}{{hr}}\]
Let us take the time taken to cover distance \[{d_1}\] while going is \[{t_1}\] and returning distance of \[{d_2}\] is \[{t_2}\] then,
\[{t_1} = \dfrac{{distance}}{{speed}} = \dfrac{{200km}}{{\dfrac{{400km}}{{hr}}}} = \dfrac{1}{2}hr\], similarly
\[{t_2} = \dfrac{{200km}}{{\dfrac{{200km}}{{hr}}}} = 1hr\],
Then the average speed is given by \[v = \dfrac{{{d_1} + {d_2}}}{{{t_1} + {t_2}}}\], where \[{d_1}\]and \[{d_2}\] is the distance while going and returning respectively.
\[ \Rightarrow v = \dfrac{{(200 + 200)km}}{{(1 + 1/2)hr}} = \dfrac{{\dfrac{{800}}{3}km}}{{hr}}\]
\[ \Rightarrow v = \dfrac{{266.6km}}{{hr}}\]
Hence the answer is option \[\left( A \right)\]$lessthan300\dfrac{{km}}{{hr}}$

Note:Speed can be converted from km/hr to m/s by using
\[1km/hr = \dfrac{5}{{18}}m/s\].
${\text{Average speed}} = \dfrac{{{\text{total distance travelled}}}}{{{\text{total time taken}}}}$\[]\] whereas in
${\text{average velocity}} = \dfrac{{{\text{displacement}}}}{{{\text{time}}}}$
In this case average velocity is zero as displacement is zero. Average speed is also given by \[\dfrac{{{v_1} + {v_2}}}{2}\], This is used only when time takes for \[{d_1}\]and \[{d_2}\] is same.