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

Solution :

$$\eqalign{ & {\text{Average}}\,{\text{acceleration}} = \frac{{{\text{change}}\,{\text{in}}\,{\text{velocity}}}}{{{\text{time interval}}}} \cr & = \frac{{\Delta \overrightarrow v }}{t} \cr & \overrightarrow {{v_1}} = 5\hat i,\overrightarrow {{v_2}} = 5\hat j \cr & \Delta \overrightarrow v = \left( {\overrightarrow {{v_2}} - \overrightarrow {{v_1}} } \right) = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos {{90}^ \circ }} \cr & = \sqrt {{5^2} + {5^2} + 0} \cr & \left[ {{\text{As}}\,\,\left| {{v_1}} \right| = \left| {{v_2}} \right| = 5\,m/s} \right] = 5\sqrt 2 \,m/s \cr & {\text{Avg}}{\text{.}}\,{\text{acc}}{\text{.}}\, = \frac{{\Delta \overrightarrow v }}{t} = \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}m/{s^2} \cr & \Rightarrow \tan \theta = \frac{5}{{ - 5}} = - 1 \cr} $$
which means $$\theta $$ is in the second quadrant. (towards north-west)