Question

A motorcyclist drives from A to B with a uniform speed of 30 km/hr and further to a point C on a
straight line with a speed of 20 km/hr. If it takes equal time in each case, then calculate its average
velocity

Answer 1

$\Huge{\underline{\underline{\mathfrak{\red{Answer :}}}}}$

A motorcyclist drives from A to B with velocity of 30 km/h , And also drives From B to C with velocity of 20 km/h. And time is Same So,

$\large \tt{Given} \begin{cases} {\sf{Velocity \: from \: A \: to \: B \: (V_{AB}) \: = \: 30 \: kmh^{-1}}} \\ {\sf{Velocity \: from \: B \: to \: C \: (V_{BC}) \: = \: 20 \: kmh^{-1}}} \\ {\sf{Time \: (t) \: = \: Constant \: (t_1 \: = \: t_2)}} \end{cases}$

If time interval is same then Acceleration will be constant because,

$\implies{\boxed{\sf{a \: = \: \frac{ (v_{1} \: - \: v _{2})}{t} }}}$

If T is constant then a = constant.

$\implies {\boxed {\boxed{\sf{acceleration \: (a) \: = \: constant }}}}$

Now use formula,

$\LARGE{\underline{\boxed{\sf{V_{avg} \: = \: \frac{V_{AB} \: + \: V_{BC}}{2}}}}}$

Put Values

⇒V avg = 20 + 30/2

⇒V avg = 50/2

⇒V avg = 25

$\huge \implies {\boxed{\sf{V_{avg} \: = \: 25 \: kmh^{-1}}}}$

∴ Average velocity is 25 km/h

Answer 2

Answer:

Given:

Cyclist travels from A to B at a speed of 30 km/hr. He again travels from B to C at a speed of 20 km/hr.

Time taken in both cases is same.

To find:

Average Velocity of the cyclist

Concept:

Always remember that average Velocity is the ratio of total distance to the total time taken.

Calculation:

Let time taken in both journeys be t

So, distance from A to B = 30t

And Distance from B to C = 20t

And total time = t + t = 2t

$avg. \: v \: = \frac{total \: distance}{total \: time}$

$= > avg. \: v = \frac{(30t + 20t)}{(2t)}$

$= > avg. \: v = \frac{50t}{2t}$

$= > avg. \: v = 25 \: km {hr}^{ - 1}$

So average Velocity is 25 km/hr.