A particle is moving eastwards with a velocity of 5\,{m}/{s}. In 10 seconds the velocity changes to 5\,{m}/{s} northwards. The average acceleration in this time is [AIEEE 2005]

Correct Answer is : 12 m s−2{1}{{2}}\,{m}\,{s}^{-2}21ms−2 towards north-west

1/2\,{m}\,{s}^{-2} towards north

{1}{{2}}\,{m}\,{s}^{-2} towards north-east

{1}{{2}}\,{m}\,{s}^{-2} towards north-west

Correct Answer is : 12 m s−2{1}{{2}}\,{m}\,{s}^{-2}21ms−2 towards north-west

Test Index

Motion in a Plane

Section: Physics

Question : 91 of 94

Marks: +1, -0

5\,\text{m}/\text{s}

5\,\text{m}/\text{s}

Solution:

Average acceleration

= \frac{\text{change in velocioty}}{\text{time interval}}

= \frac{\overset{\rightarrow}{\Delta v}}{t}

\overset{\rightarrow}{v_1}=+5\hat{i}

, towards east direction

\overset{\rightarrow}{v_2}=+5\hat{j}

, towards north direction

\;\Delta \overset{\rightarrow}{v}=\overset{\rightarrow}{v_2}-\overset{\rightarrow}{v_1}=5\hat{i}-5\hat{j}

\;\therefore \overset{\rightarrow}{a}=\frac{\widehat{5j-5\hat{i}}}{10}=\frac{\widehat{j-\hat{i}}}{2}

\;\therefore \;\; a=\frac{\sqrt{1^2+(-1)^2}}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\,\text{m}\,\text{s}^{-2}

\;\tan\theta=\frac{v_2}{v_1}=\frac{5}{5}=1

\;\therefore \;\; \theta=45^{\circ}

Therefore the direction is North-west.

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