Question

a car moving on astraight road covers one third dustance with 20km per h and the rest with 60km per h.the average speed of car is ​

Answer 1

Solution:

➡ Given:

✏ 1/3 of total distance is covered by a car with velocity equal to 20kmph.

✏ The rest distance is covered with velocity equal to 60kmph.

➡ To Find:

✏ The average speed of car.

➡ Concept:

✏ Average speed is defined as total distance covered by a body in total time.

➡ Formula:

✏ Formula of average speed is given by

$\star \: \underline{ \boxed{ \bold{ \rm{ \pink{v_{av} = \frac{total \: distance}{total \: time}}}}}}$

➡ Calculation:

✏ Let body covers 1/3 of total distance in time t1 and 2/3 of total distance in time t2.

✏ Total distance covered by body is 'd'.

$\implies \rm \: v_{av} = \frac{d}{t_1 + t_2} \\ \\ \implies \rm \: v_{av} = \frac{d}{ \frac{d}{3v_{1}} + \frac{2d}{3v_{2}} } \: \: ( \because{t} = \frac{d}{v} ) \\ \\ \implies \rm \: v_{av} = \frac{ \cancel{d} \times 3v_1v_2}{ \cancel{d}(v_{2} + 2v_{1})} \\ \\ \implies\rm \: v_{av} = \frac{3 \times 20 \times 60}{60 + (2 \times 20)} \\ \\ \implies \rm \: v_{av} = \frac{3600}{100} \\ \\ \implies \: \underline{ \boxed{ \bold{ \rm{ \red{v_{av} = 36 \: kmph}}}}} \: \star$

Answer 2

✏ 1/3 of total distance is covered by a car with velocity equal to 20kmph.

✏ The rest distance is covered with velocity equal to 60kmph.

➡ To Find:

✏ The average speed of car.

➡ Concept:

✏ Average speed is defined as total distance covered by a body in total time.

➡ Formula:

✏ Formula of average speed is given by

\star \: \underline{ \boxed{ \bold{ \rm{ \pink{v_{av} = \frac{total \: distance}{total \: time}}}}}}⋆

v

av

=

totaltime

totaldistance

➡ Calculation:

✏ Let body covers 1/3 of total distance in time t1 and 2/3 of total distance in time t2.

✏ Total distance covered by body is 'd'.

\begin{lgathered}\implies \rm \: v_{av} = \frac{d}{t_1 + t_2} \\ \\ \implies \rm \: v_{av} = \frac{d}{ \frac{d}{3v_{1}} + \frac{2d}{3v_{2}} } \: \: ( \because{t} = \frac{d}{v} ) \\ \\ \implies \rm \: v_{av} = \frac{ \cancel{d} \times 3v_1v_2}{ \cancel{d}(v_{2} + 2v_{1})} \\ \\ \implies\rm \: v_{av} = \frac{3 \times 20 \times 60}{60 + (2 \times 20)} \\ \\ \implies \rm \: v_{av} = \frac{3600}{100} \\ \\ \implies \: \underline{ \boxed{ \bold{ \rm{ \red{v_{av} = 36 \: kmph}}}}} \: \star\end{lgathered}

⟹v

av

=

t

1

+t

2

d

⟹v

av

=

3v

1

d

+

3v

2

2d

d

(∵t=

v

d

)

⟹v

av

=

d

(v

2

+2v

1

)

d

×3v

1

v

2

⟹v

av

=

60+(2×20)

3×20×60

⟹v

av

=

100

3600

⟹

v

av

=36kmph

hope it helps you.....