Question

a cyclist travels a distance of 1.4 km in the first hour 0.25 km in the second hour and 0.15 km in the third hour find the average speed in metre per second​

Answer 1

Given :

• Distance travlled by the cyclist in first hour = 1.4 km
• Distance travelled by the cyclist in second hour = 0.25 km
• Distance travelled by the cyclist in third hour = 0.15 km

To Find :

• The average speed of the cyclist in metres per second

Solution :

Total distance travelled by the cyclist = 1.4 km + 0.25 km + 0.15 km = 1.8 km = 1.8 × 1000 m = 1800 m

Total time taken = 3 hours = 3 × 3600 sec = 10,800 sec

Average speed is calculated by the ratio between total distance travelled to total time taken.

We have ,

• Total distance travelled = 1800 m
• Total time taken = 10,800 sec

➙ Average speed = 1800 m/ 10800 sec

➙ Average speed = 0.16 m/s

Hence ,

• The average speed of the cyclist in metres per sec is 0.16 m/s

Answer 2

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

• A cyclist travels a distance of 1.4 km in 1st hour.

• And 0.25 km in 2nd hour.

• And 0.15 km in 3rd hour.

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

• The average speed.

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

$\bf\red{According\:to\:question,}$

• $\bf{d_1\:=\:1.4\:km}$

• $\bf{d_2\:=\:0.25\:km}$

• $\bf{d_3\:=\:0.15\:km}$

$\longmapsto\:\:\bf{Total\:distance\:(d)\:=\:d_1\:+\:d_2\:+\:d_3\:}$

$\longmapsto\:\:\bf{Total\:distance\:(d)\:=\:1.4\:+\:0.25\:+\:0.15\:}$

$\longmapsto\:\:\bf\blue{Total\:distance\:(d)\:=\:1.8\:km.\:}$

$\longmapsto\:\:\bf\purple{Total\:time\:(t)\:=\:3\:hr.\:}$

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

$\orange\bigstar\:\:\bf\green{Average\:velocity\:(A_s)\:=\:\dfrac{Total\:distance}{Total\:time}}$

$:\implies\:\:\bf{A_s\:=\:\dfrac{1.8}{3}\:}$

$:\implies\:\:\bf{\color{olive}A_s\:=\:0.6\:km/hr\:}$

$:\implies\:\:\bf{A_s\:=\:0.6\times{\dfrac{1000}{3600}}\:m/s\:}$

$:\implies\:\:\bf{A_s\:=\:0.6\times{\dfrac{10}{36}}\:m/s\:}$

$:\implies\:\:\bf{A_s\:=\:0.6\times{\dfrac{5}{18}}\:m/s\:}$

$:\implies\:\:\bf{A_s\:=\:\dfrac{3}{18}\:m/s\:}$

$:\implies\:\:\bf{A_s\:=\:\dfrac{1}{6}\:m/s\:}$

$:\implies\:\:\bf\orange{A_s\:=\:0.167\:m/s\:}$

$\Large\bold\therefore$ Average speed of the cyclist is 0.167m/s.