An object moves $8 m$ East and then $6 m$ North in $10 s$ . Find the average speed and average velocity of the object respectively.

$1.5 m / s, 1.4 m / s$

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$1.4 m / s, 1.4 m / s$

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$1.4 m / s, 1 m / s$

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None of these

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Solution

The correct option is C
$1.4 m / s, 1 m / s$

Distance covered by the object:
$d = A B + B C = 8 + 6 = 14 m$

Applying Pythagoras theorem in $△ A B C$
$A C^{2} = A B^{2} + B C^{2}$
$A C = \sqrt{A B^{2} + B C^{2}}$
$A C = \sqrt{8^{2} + 6^{2}}$
$A C = \sqrt{100} = 10 m$
Displacement of the object: $s = A C = 10 m$

Time: $t = 10 s$

$Average speed = \frac{Distance}{time}$
$Average speed = \frac{14}{10} = 1.4 m / s$

$Average velocity = \frac{Displacement}{time}$
$Average velocity = \frac{10}{10} = 1 m / s$

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