Question

A car moves a distance of 200m . it covers the first half of the distance at speed 60kmperhr and second half of the distance at speed v . the average speed is 40kmperhr . the value of v is ?​

Answer 1

Question :-

A car moves a distance of 200m ,if it covers the first half of the distance at speed of 60km/hr and second half of the distance at speed v , the average speed is 40km/hr . Then ,find the value of v .

To Find :-

The value of v or the speed of the car to cover the second half of the journey.

Given :-

• Distance (s) = 200 km

• Speed of the car in the first half = 60 km/hr

• Speed of the car in second half = v km/hr

• Average Speed = 40 km/hr

We know :-

Formula For AveraGe SPeed :-

$\boxed{\underline{\bf{Average\:Speed = \dfrac{s_{1} + s_{2}}{t_{1} + t_{2}}}}}$

Where :-

• s = Distance Covered
• t = Time Taken

Fornula For SPeed :-

$\boxed{\underline{\bf{Speed = \dfrac{s}{t}}}}$

Where :-

• s = Distance Covered
• t = Time Taken

Concept :-

According to the Question, the Car First half of the distance in 60 km/hr .i.e,

The car travels 100 km in 60 km/hr.

For the second half the car covers the distance in v km/hr. i.e,

It covers 100km in v km/hr.

Now by this information we can Find the time Taken of the two journey's .

After finding the time taken we can find the the speed of the car in the second part.

Solution :-

To Find the time taken for the First Half :-

• Speed$\bf{(v_{1}) = 60 km h^{-1}}$.

• Distance = 100 km

[Note : The Distance is taken as 100 km , since in the question it said that the car cover half of the distance]

Let the time taken be $\bf{t_{1}}$.

Using the formula for speed and substituting the values in it , we get :-

$:\implies \bf{v = \dfrac{s}{t}} \\ \\ \\ :\implies \bf{v_{1} = \dfrac{s_{1}}{t_{1}}} \\ \\ \\ :\implies \bf{60 = \dfrac{100}{t_{1}}} \\ \\ \\ :\implies \bf{60 \times t_{1} = 100} \\ \\ \\ :\implies \bf{t_{1} = \dfrac{100}{60}} \\ \\ \\ :\implies \bf{t_{1} = \dfrac{10\not{0}}{6\not{0}}} \\ \\ \\ :\implies \bf{t_{1} = \dfrac{10}{6}} \\ \\ \\ \therefore \purple{\bf{t_{1} = \dfrac{10}{6}}}$

Hence, the time taken is 10/6 hr.

To Find the time taken for the Second Half :-

• Speed$\bf{(v_{2}) = v km h^{-1}}$.

• Distance = 100 km

[Note : The Distance is taken as 100 km , since in the question it said that the car cover half of the distance]

Let the time taken be $\bf{t_{2}}$.

Using the formula for speed and substituting the values in it , we get :-

$:\implies \bf{v = \dfrac{s}{t}} \\ \\ \\ :\implies \bf{v_{2} = \dfrac{s_{2}}{t_{2}}} \\ \\ \\ :\implies \bf{v = \dfrac{100}{t_{2}}} \\ \\ \\ :\implies \bf{v_{2} \times t_{2} = 100} \\ \\ \\ :\implies \bf{t_{2} = \dfrac{100}{v_{2}}} \\ \\ \\ :\implies \bf{t_{1} = \dfrac{100}{v_{2}}} \\ \\ \\ \therefore \purple{\bf{t_{1} = \dfrac{100}{v_{2}}}}$

Hence, the time taken is 100/v hr.

Hence, the time taken is 100/v hr.

Speed of the car to cover Second-half :-

Given :-

• Total Distance = 200 km

• Time$\bf{(t_{1}) = \dfrac{10}{6}h}$

• Time$\bf{(t_{2}) = \dfrac{100}{v_{2}}h}$

Using the Formula for average Speed and substituting the values in it , we get :-

$:\implies \bf{Average\:Speed = \dfrac{s_{1} + s_{2}}{t_{1} + t_{2}}} \\ \\ \\ \\ :\implies \bf{40 = \dfrac{200}{\dfrac{10}{6} + \dfrac{100}{v_{2}}}} \:\: [s_{1} + s_{2} = 200 km]\\ \\ \\ \\ :\implies \bf{40 = \dfrac{200}{\dfrac{10v_{2} + 1200}{6v_{2}}}} \\ \\ \\ \\ :\implies \bf{40 = \dfrac{200}{10v_{2} + 1200} \times 6v_{2}}$

$:\implies \bf{40(10v_{2} + 600) = 1200v_{2}} \\ \\ \\ \\ :\implies \bf{400v_{2} + 24000 = 1200v_{2}} \\ \\ \\ \\ :\implies \bf{24000 = 1200v_{2} - 4000v_{2}} \\ \\ \\ \\ :\implies \bf{24000 = 800v_{2}} \\ \\ \\ \\ :\implies \bf{\dfrac{24000}{800} = v_{2}} \\ \\ \\ \\ :\implies \bf{30 = v_{2}} \\ \\ \\ \\ \therefore \purple{v_{2} = 30 km h^{-1}}$

Hence, the speed taken to cover the second half is 30 km/h.