Question

A cyclist travels a distance of 4 kilometre from a to b and then moves a distance of 3 km right angle to a b find his displacement and distance travelled

Answer 1

Answer:

• The Displacement of the cyclist is 5 Km.
• The Distance of the cyclist is 7 Km.

Given:

1. A B Length = 4 Km.
2. B C Length = 3 Km.

Explanation:

$\rule{300}{1.5}$

#Refer the attachment for figure.

Displacement:

It is a Shortest path between Initial and Final position of the body.

Now, in the given in Figure,

The shortest path is A C

Applying, Pythagoras theorem,

⇒ AC² = AB² + BC²

Substituting the values,

⇒ AC² = (4)² + (3)²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = √ 25

⇒ AC = 5

⇒AC = 5 Km

⇒ Displacement = AC = 5 Km.

∴ The Displacement of the cyclist is 5 Km.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Distance:

The total length of the path covered by the body during its travel.

Now, in the given figure,

Distance is sum of AB and BC

⇒ Distance = AB + BC

Substituting the values,

⇒ Distance = 4 Km + 3 Km

⇒ Distance = 4 + 3

⇒ Distance = 7

⇒ Distance = 7 Km

∴ The Distance of the cyclist is 7 Km.

$\rule{300}{1.5}$

Answer 2

[tex]\huge\underline\mathtt\red{Answer:-}[/tex

•The Displacement of the cyclist is 5 Km.

•The Displacement of the cyclist is 5 Km.•The Distance of the cyclist is 7 Km.