A particle velocity changes from (2i^-3j^)ms-1to (3i^-2j^)ms-1 in 2 s. If its mass is 1 kg, the acceleration (ms-2) is

a=vf-vit=(3i^-2i^)-(2j^-3j^)2=i^+j^2

A particle velocity changes from (2i^-3j^)ms-1to (3i^-2j^)ms-1 in 2 s. If its mass is 1 kg, the acceleration (ms-2) is

a=vf-vit=(3i^-2i^)-(2j^-3j^)2=i^+j^2

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A particle velocity changes from $(2 \hat{i} - 3 \hat{j}) {ms}^{- 1}$ to $(3 \hat{i} - 2 \hat{j}) {ms}^{- 1}$ in 2 s. If its mass is 1 kg, the acceleration ( $m s^{- 2}$ ) is

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zero

$- (\hat{i} + \hat{j})$

$(\hat{i} - \hat{j}) / 2$

$(\hat{i} + \hat{j}) / 2$

answer is B.

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Detailed Solution

$a = \frac{v_{f} - v_{i}}{t} = \frac{(3 \hat{i} - 2 \hat{i}) - (2 \hat{j} - 3 \hat{j})}{2} = \frac{\hat{i} + \hat{j}}{2}$

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A particle velocity changes from (2i^-3j^)ms-1to (3i^-2j^)ms-1 in 2 s. If its mass is 1 kg, the acceleration (ms-2) is

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A particle velocity changes from $(2 \hat{i} - 3 \hat{j}) {ms}^{- 1}$ to $(3 \hat{i} - 2 \hat{j}) {ms}^{- 1}$ in 2 s. If its mass is 1 kg, the acceleration ( $m s^{- 2}$ ) is