Question

A car moves at a speed of \[60\,kmh{r^{ - 1}}\] from point A to point B and then with the speed of \[40\,kmh{r^{ - 1}}\] from point B to point C. Further, it moves to point D with a speed equal to its average speed between A and C. Points A. B, C and D are collinear and equidistant. Find the average speed of the car between A and D.
A. \[30\,kmh{r^{ - 1}}\]
B. \[50\,kmh{r^{ - 1}}\]
C. \[48\,kmh{r^{ - 1}}\]
D. \[60\,kmh{r^{ - 1}}\]

Answer 1

Hint: The ratio of total distance travelled to total time travelled yields the average velocity of a particle. As a result, we can calculate the total distance covered by a particle as the product of its velocity and the time it takes to move from one location to another.

Formula Used:
To find the average speed of a particle the formula is,
\[{V_{avg}} = \dfrac{x}{t}\]
Where, \[x\] is the total distance travelled and \[t\] is the total time taken.

Complete step by step solution:
Consider a car which moves with a speed of \[60kmh{r^{ - 1}}\]from point A to point B and then with the speed of \[40kmh{r^{ - 1}}\]A and C. The points A, B, C, and D are collinear and equidistant. We need to find the average speed of the car between A and D. The car moves from A to B at \[60\,kmh{r^{ - 1}}\], so the car can cover 60 km in 1hr. Therefore, to cover AB distance, time taken is,
\[{t_1} = \dfrac{{AB}}{{60}}\]

Similarly, the car moves from B to C at \[40\,kmh{r^{ - 1}}\], so the car can cover 40km in 1hr. Therefore, to cover AC distance, time taken is,
\[{t_2} = \dfrac{{BC}}{{60}}\]
Since it is given that AB and BC are equidistant and collinear. Therefore, \[AB = BC\], so replace either AB with BC or BC with AB.

The average speed is,
\[{V_{avg}} = \dfrac{x}{t}\]
\[\Rightarrow {V_{avg}} = \dfrac{{AB + BC}}{{\left( {\dfrac{{AB}}{{60}} + \dfrac{{BC}}{{40}}} \right)}} \\ \]
\[\Rightarrow {V_{avg}} = \dfrac{{2AB}}{{\left( {\dfrac{{AB}}{{60}} + \dfrac{{AB}}{{40}}} \right)}} \\ \]
\[\Rightarrow {V_{avg}} = \dfrac{{2AB}}{{\left( {\dfrac{{5AB}}{{120}}} \right)}} \\ \]
\[\therefore {V_{avg}} = 48\,kmh{r^{ - 1}}\]
As the average speed between A and C is equal to the speed for CD so the answer is \[48\,kmh{r^{ - 1}}\]. Therefore, the average speed of the car between A and D is \[48\,kmh{r^{ - 1}}\].

Hence, option C is the correct answer.

Note:Remember that, since the points A, B, C and D are collinear and equidistant, that is the distance between A to B, B to C and C to D is the same. Therefore, we can replace AB and BC as shown in this solution.