Question

EXAMPLE
A
A body moving along a straight path covers one-third of a
distance with a velocity of 5 Kmph and the rest of the path
with a velocity of 20 Kmph. Find the average velocity of the
body.
A X/
2x/3
с
X​

Answer 1

Explanation:

As per the provided information in the given question, we have :

• A body moving along a straight path covers one-third of distance with a velocity of 5 km/h.
• And, the rest of the path with a velocity of 20 km/h.

We are asked to calculate the average velocity.

$\longmapsto \small \bf {Velocity_{(avg)} = \dfrac{Total \; displacement}{Total\; time} }$

Calculating total displacement :

Firstly, let us suppose total distance travelled as x. Since, the body is moving along a straight path in a single direction. So, total displacement will be equal to the total distance travelled.

⇒ Total displacement = x km

Calculating total time :

• The body covers one third of the distance with the velocity of 5 km/h. (Here, distance can be considered as displacement as the body is moving along a straight path in a single direction.)

$\longmapsto \rm {t_1 = \dfrac{s_1}{v_1} }$

• t denotes time
• s denotes displacement
• v denotes velocity

$\longmapsto \rm {t_1 = \Bigg ( \dfrac{1}{3}x \div 5 \Bigg ) \; h }$

$\longmapsto \rm {t_1 = \Bigg ( \dfrac{1}{3}x \times \dfrac{1}{5} \Bigg ) \; h }$

$\longmapsto \rm {t_1 = \dfrac{x}{15} \; h }$

Similarly,

• The body covers rest distance with the velocity of 20 km/h. (Here, distance can be considered as displacement as the body is moving along a straight path in a single direction.)

$\longmapsto \rm {t_2 = \dfrac{s_2}{v_2} }$

$\longmapsto \rm {t_2 = \Bigg \lgroup \Bigg ( x - \dfrac{1}{3}x \Bigg ) \div 20 \Bigg \rgroup \; h }$

$\longmapsto \rm {t_2 = \Bigg \lgroup \Bigg ( \dfrac{3x -x}{3}\Bigg ) \div 20 \Bigg \rgroup \; h }$

$\longmapsto \rm {t_2 = \Bigg \lgroup \dfrac{2x}{3} \div 20 \Bigg \rgroup \; h }$

$\longmapsto \rm {t_2 = \Bigg \lgroup \dfrac{2x}{3} \times \dfrac{1}{20} \Bigg \rgroup \; h }$

$\longmapsto \rm {t_2 = \Bigg \lgroup \dfrac{x}{3} \times \dfrac{1}{10} \Bigg \rgroup \; h }$

$\longmapsto \rm {t_2 = \dfrac{x}{30} \; h }$

Therefore,

$\longmapsto \rm { Time_{(Total)} = t_1 + t_2 }$

$\longmapsto \rm { Time_{(Total)} = \Bigg ( \dfrac{x}{15} + \dfrac{x}{30} \Bigg ) \; h }$

$\longmapsto \rm { Time_{(Total)} = \Bigg ( \dfrac{2x + x}{30} \Bigg ) \; h }$

$\longmapsto \rm { Time_{(Total)} = \Bigg ( \dfrac{3x}{30} \Bigg ) \; h }$

$\longmapsto \bf { Time_{(Total)} = \dfrac{x}{10} \; h }$

Calculating average velocity :

$\longmapsto \rm {Velocity_{(avg)} = \dfrac{Total \; displacement}{Total\; time} }$

$\longmapsto \rm {Velocity_{(avg)} = \Bigg ( x \div \dfrac{x}{10} \Bigg ) \; kmh^{-1} }$

$\longmapsto \rm {Velocity_{(avg)} = \Bigg ( x \times \dfrac{10}{x} \Bigg ) \; kmh^{-1} }$

$\longmapsto \bf {Velocity_{(avg)} = 10 \; kmh^{-1} }$

∴ Magnitude of average velocity is 10 km/h.

Answer 2

Let the total distance be = x

Given that,

x/3 of x is travelled with a speed of 5 km/h.

And, 2x/3 is travelled with a speed of 20 km/h.

{Rest = x - x/3.}

So, t1 = x1/v1 = (x/3)/(5) = x/15

t2 = x2/v2 = (2x/3)/(20) = x/30

Now we know,

$\sf \overline{v} = \dfrac{Total \ distance}{Total \ time}$

⇒ v = (x)/(x/15 + x/30)

⇒ v = (1)/(2 + 1/30)

⇒ v = 30/3 = 10 km/h.

Therefore, average speed was of 10 km/h.