Question

A car travels 100 km at a speed of 60 km/h and returns with a speed of 40 km/h . calculate its average speed and average velocity​

Answer 1

Answer :-

Given :-

A car travels 100 km at a speed of 60 km/hr .

Returns back with a speed of 40 km/hr .

Required to find :-

• Average Speed ?

• Average velocity ?

Concept used :-

1. Speed and velocity

Formulae used :-

$\large{\leadsto{\boxed{\rm{Average \; speed = \dfrac{Total \; distance }{Total \; time \; taken }}}}}$

$\large{\leadsto{\boxed{\rm{Average \; velocity = \dfrac{Total \; displacement }{Total \; time \; taken }}}}}$

Solution :-

Given that :-

A car travels 100 km at a speed of 60 km/hr .

Returns back with a speed of 40 km/hr

He asked us to find the average speed and velocity .

So, Let's consider this as 2 cases

Case 1 ( while going )

Case 2 ( while returning back )

So,

Case - 1

Distance ( d1) = 100 km

Speed (S1) = 60 km/hr

Time ( t1) = ?

we know that ,

$\rm{ speed = \dfrac{Distance}{time}}$

So,

$\rm{ Time = \dfrac{Distance}{Speed}}$

$\longrightarrow{\tt{ Time = \dfrac{ 100}{60}}}$

$\longrightarrow{\tt{ Time ({t}_{1}) = \dfrac{5}{3}\;hours}}$

Similarly,

Case - 2

Distance ( d2 ) = 100 km

Speed ( S2 ) = 40 km/hr

Time ( t2 ) = ?

Hence,

$\longrightarrow{\tt{ Time = \dfrac{ 100 }{40 }}}$

$\longrightarrow{\tt{ Time \; ( {t}_{2} ) = \dfrac{5}{2} \; hours }}$

Now,

From case 1 and case 2 we can conclude that ,

Total distance = d1 + d2

=> 100 km + 100 km

=> 200 km

Total displacement = 0

Total time taken = t1 + t2

=> $\rm{ \dfrac{5}{3} + \dfrac{5}{2}}$

=> $\rm{ \dfrac{ 10 + 15 }{6}}$

=> $\rm{ \dfrac{25}{6}\;hours}$

Hence , using the formula

$\large{\leadsto{\boxed{\rm{Average \; speed = \dfrac{Total \; distance }{Total \; time \; taken }}}}}$

$\longrightarrow{\tt{ Average \; speed = \dfrac{ 200 \; km }{ \dfrac{25}{6}\; hr }}}$

$\longrightarrow{\tt{ Average \; speed = \dfrac{ 200 \; km \times 6 }{ 25\; hr }}}$

$\longrightarrow{\tt{ Average \; speed = 8 \times 6 }}$

$\longrightarrow{\tt{ Average \; speed = 48 \; km/hr }}$

So,

$\longrightarrow{\boxed{\tt{ Average \; speed = 48 \; km/hr }}}$

Similarly,

Using the formula ;

$\large{\leadsto{\boxed{\rm{Average \; velocity = \dfrac{Total \; displacement }{Total \; time \; taken }}}}}$

$\longrightarrow{\tt{ Average \; velocity = \dfrac{ 0 \; km }{ \dfrac{25}{6} \; hr }}}$

$\longrightarrow{\tt{ Average \; velocity = 0 \; km/hr }}$

Point to remember :

1. If a body travels a distance from a point and again reaches the same point after traveling some distance then the displacement is zero . so that is why total displacement is zero .

Answer 2

Answer:

Answer :-

Given :-

A car travels 100 km at a speed of 60 km/hr .

Returns back with a speed of 40 km/hr .

Required to find :-

Average Speed ?

Average velocity ?

Concept used :-

Speed and velocity

Formulae used :-

\large{\leadsto{\boxed{\rm{Average \; speed = \dfrac{Total \; distance }{Total \; time \; taken }}}}}⇝

Averagespeed=

Totaltimetaken

Totaldistance

\large{\leadsto{\boxed{\rm{Average \; velocity = \dfrac{Total \; displacement }{Total \; time \; taken }}}}}⇝

Averagevelocity=

Totaltimetaken

Totaldisplacement

Solution :-

Given that :-

A car travels 100 km at a speed of 60 km/hr .

Returns back with a speed of 40 km/hr

He asked us to find the average speed and velocity .

So, Let's consider this as 2 cases

Case 1 ( while going )

Case 2 ( while returning back )

So,

Case - 1

Distance ( d1) = 100 km

Speed (S1) = 60 km/hr

Time ( t1) = ?

we know that ,

\rm{ speed = \dfrac{Distance}{time}}speed=

time

Distance

So,

\rm{ Time = \dfrac{Distance}{Speed}}Time=

Speed

Distance

\longrightarrow{\tt{ Time = \dfrac{ 100}{60}}}⟶Time=

60

100

\longrightarrow{\tt{ Time ({t}_{1}) = \dfrac{5}{3}\;hours}}⟶Time(t

1

)=

3

5

hours

Similarly,

Case - 2

Distance ( d2 ) = 100 km

Speed ( S2 ) = 40 km/hr

Time ( t2 ) = ?

Hence,

\longrightarrow{\tt{ Time = \dfrac{ 100 }{40 }}}⟶Time=

40

100

\longrightarrow{\tt{ Time \; ( {t}_{2} ) = \dfrac{5}{2} \; hours }}⟶Time(t

2

)=

2

5

hours

Now,

From case 1 and case 2 we can conclude that ,

Total distance = d1 + d2

=> 100 km + 100 km

=> 200 km

Total displacement = 0

Total time taken = t1 + t2

=> \rm{ \dfrac{5}{3} + \dfrac{5}{2}}

3

5

+

2

5

=> \rm{ \dfrac{ 10 + 15 }{6}}

6

10+15

=> \rm{ \dfrac{25}{6}\;hours}

6

25

hours

Hence , using the formula

\large{\leadsto{\boxed{\rm{Average \; speed = \dfrac{Total \; distance }{Total \; time \; taken }}}}}⇝

Averagespeed=

Totaltimetaken

Totaldistance

\longrightarrow{\tt{ Average \; speed = \dfrac{ 200 \; km }{ \dfrac{25}{6}\; hr }}}⟶Averagespeed=

6

25

hr

200km

\longrightarrow{\tt{ Average \; speed = \dfrac{ 200 \; km \times 6 }{ 25\; hr }}}⟶Averagespeed=

25hr

200km×6

\longrightarrow{\tt{ Average \; speed = 8 \times 6 }}⟶Averagespeed=8×6

\longrightarrow{\tt{ Average \; speed = 48 \; km/hr }}⟶Averagespeed=48km/hr

So,

\longrightarrow{\boxed{\tt{ Average \; speed = 48 \; km/hr }}}⟶

Averagespeed=48km/hr

Similarly,

Using the formula ;

\large{\leadsto{\boxed{\rm{Average \; velocity = \dfrac{Total \; displacement }{Total \; time \; taken }}}}}⇝

Averagevelocity=

Totaltimetaken

Totaldisplacement

\longrightarrow{\tt{ Average \; velocity = \dfrac{ 0 \; km }{ \dfrac{25}{6} \; hr }}}⟶Averagevelocity=

6

25

hr

0km

\longrightarrow{\tt{ Average \; velocity = 0 \; km/hr }}⟶Averagevelocity=0km/hr

Point to remember :

1. If a body travels a distance from a point and again reaches the same point after traveling some distance then the displacement is zero . so that is why total displacement is zero .