Question

The train travels with a speed of $60km{H^{ - 1}}$ from station A to station B and then comes back with a speed $80km{H^{ - 1}}$ from station B to station A. Find The average velocity of train.

Answer 1

Hint : Use the definition of velocity to solve the problem. The velocity of an object is the displacement covered per time. $\vec v = \dfrac{{\vec s}}{t}$ . Where, $\vec v$ is the velocity of the object and $\vec s$ is the displacement of the object.

Complete Step By Step Answer:
Here, we have given that the train goes from station A to station B with a speed of $60km{H^{ - 1}}$ and comes back from B to A with a speed of $80km{H^{ - 1}}$ . Now, we know the average velocity of an object is the net displacement of the object by total time taken to cover the displacement.
Now, let say the distance between the objects is $xkm$ .
So, time taken to traverse from station A to station B with a speed of $60km{H^{ - 1}}$ will be, $\dfrac{x}{{60}}H$
and time taken to traverse from B to A with a speed of $80km{H^{ - 1}}$ will be , $\dfrac{x}{{80}}H$ .
Hence, displacement of the train by going from station A to station B will be $\vec xkm$ . Also, displacement of the train from coming back to A from be will be $- \vec xkm$ .
Hence, net displacement will become, $\vec s = (\vec x - \vec x)km = 0km$ .
Hence, average velocity of the train will be,
$\vec v = \dfrac{{\vec s}}{t} = \dfrac{0}{{\dfrac{x}{{60}} + \dfrac{x}{{80}}}} = 0$ [Where, $t$ is the total time taken to traverse from A to B And B to A ]
Hence, the average velocity of the train will be $0km{H^{ - 1}}$ .

Note :
Average speed of an object is given by, the total distance traversed by total time taken by the object. If we calculate the average speed of the train here, that will not be equal to zero. As, distance cannot be negative since it is a scalar quantity whereas displacement is a vector quantity.