Question

A train travels at a speed 50 km/hr for 0.5 hr, at 30
km/hr for next 0.26 hr and then 70 km/hr for next
0.76 hr. What is the average speed of the train?​

Answer 1

Given :-

Speed of the train in 0.5 h = 50 km/h

Speed of the train in 0.26 h = 30 km/h

Speed of the train in 0.76 h = 70 km/h

To Find :-

The average speed of the train.

Analysis :-

Firstly add up all the distance and time taken by the train.

In order to find the average speed substitute the values we got accordingly such that total distance divided by the total time taken.

Solution :-

We know that,

• s = Speed
• d = Distance
• t = Time

Total distance,

d₁ + d₂ + d₃

Given that,

d₁ = 50 × 0.5 = 25 km

d₂ = 30 × 0.26 = 7.8 km

d₃ = 70 × 0.76 = 53.2 km

Substituting their values,

25 + 7.8 + 53.2

= 86 km

Total time,

t₁ + t₂ + t₃

Given that,

t₁ = 0.5 hr

t₂ = 0.26 hr

t₃ = 0.76 hr

Substituting their values,

0.5 + 0.26 + 0.75

= 1.52 hours

Using the formula,

$\underline{\boxed{\sf Average \ speed=\dfrac{Total \ distance}{Total \ time} }}$

Given that,

Distance (d) = 86 km

Time (t) = 1.52 hours

Substituting their values,

⇒ s = 86/1.52

⇒ s = 56.57 km/h

Therefore, the average speed of the train is 56.57 km/h.

Answer 2

$\Large\bf{\color{indigo}GiVeN,}$

• A train travels,

1. At speed 50 km/hr for 0.5 hr.
2. At speed 30 km/hr for 0.26 hr.
3. At speed 70 km/hr for 0.76 hr.

$\Large\bf{\color{coral}To\:FiNd,}$

• Average Speed of the train.

$\Large\bf{\color{cyan}CaLcUlAtIoN,}$

CASE - 1 ;-

$\bf\red{Given\:that,}$

• $\bf{v_1\:=\:50\:km/hr}$

• $\bf{t_1\:=\:0.5\:hr}$

$\bf\pink{We\:know\:that,}$

$\orange\bigstar\:\:\bf{\color{olive}Velocity\:=\:\dfrac{Distance}{Time}\:}$

$:\longmapsto\:\:\bf{Distance\:=\:Velocity\times{time}\:}$

$:\longmapsto\:\:\bf{Distance\:=\:50\times{0.5}\:}$

$:\longmapsto\:\:\bf\blue{Distance\:(D_1)\:=\:25\:km\:}$

CASE - 2 ;-

$\bf\red{Given\:that,}$

• $\bf{v_2\:=\:30\:km/hr}$

• $\bf{t_2\:=\:0.26\:hr}$

$\bf\pink{We\:know\:that,}$

$\orange\bigstar\:\:\bf{\color{olive}Velocity\:=\:\dfrac{Distance}{Time}\:}$

$:\longmapsto\:\:\bf{Distance\:=\:Velocity\times{time}\:}$

$:\longmapsto\:\:\bf{Distance\:=\:30\times{0.26}\:}$

$:\longmapsto\:\:\bf\blue{Distance\:(D_2)\:=\:7.8\:km\:}$

CASE - 3 ;-

$\bf\red{Given\:that,}$

• $\bf{v_3\:=\:70\:km/hr}$

• $\bf{t_3\:=\:0.76\:hr}$

$\bf\pink{We\:know\:that,}$

$\orange\bigstar\:\:\bf{\color{olive}Velocity\:=\:\dfrac{Distance}{Time}\:}$

$:\longmapsto\:\:\bf{Distance\:=\:Velocity\times{time}\:}$

$:\longmapsto\:\:\bf{Distance\:=\:70\times{0.76}\:}$

$:\longmapsto\:\:\bf\blue{Distance\:(D_3)\:=\:53.2\:km\:}$

__________________________

$\bf{We\:know\:that,}$

$\red\bigstar\:\:\bf{\color{purple}Average\:Speed\:=\:\dfrac{Total\:Distance}{Total\:time}\:}$

$:\implies\:\:\bf{Average\:Speed\:=\:\dfrac{D_1\:+\:D_2\:+\:D_3}{t_1\:+\:t_2\:+\:t_3}\:}$

$:\implies\:\:\bf{Average\:Speed\:=\:\dfrac{25\:+\:7.8\:+\:53.2}{0.5\:+\:0.26\:+\:0.76}\:}$

$:\implies\:\:\bf{Average\:Speed\:=\:\dfrac{86}{1.52}\:}$

$:\implies\:\:\bf\green{Average\:Speed\:=\:56.57\:km/hr\:}$

$\Large\bold\therefore$ The average speed of the train is 56.57 km/hr.