A particle is moving Eastwards with a velocity of $5 m s^{- 1} .$ In 10 s, the velocity changes to $5 m s^{- 1}$ Northwards. The average acceleration in this time is

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zero

$\frac{1}{\sqrt{2}} m s^{- 2}$ towards North-East

$\frac{1}{\sqrt{2}} m s^{- 2}$ towards North-West

$\frac{1}{2} m s^{- 2}$ towards North

answer is D.

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Detailed Solution

As, velocity of Eastward $v_{1} = + 5 \hat{i}$
Question Image
And velocity of Northward $v_{2} = + 5 \hat{j}$
$Δ v = v_{2} - v_{1} = 5 \hat{j} - 5 \hat{i}$
$| Δ v | = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2}$
$∴ a = \frac{| Δ v |}{t} = \frac{5 \sqrt{2}}{10} = \frac{1}{\sqrt{2}} m s^{- 2}$
For direction, $\tan α = - \frac{5}{5} = - 1$
Average acceleration is $\frac{1}{\sqrt{2}} m s^{- 2}$ towards North-West.