A particle is moving with velocity 5 m/s towards the east and its velocity changes to 5 m/s north in 10 sec. Find the acceleration.
(A) $\sqrt{2} m/s^2 $ N-W
(B) $\dfrac{1}{\sqrt{2}} m/s^2 $ N-W
(C) $\dfrac{1}{\sqrt{2}} m/s^2 $ N-E
(D) $\sqrt{2} m/s^2 $ N-E

Hint: The velocity initially was only along the eastward direction, later it also had components along the north direction with the same magnitude. There is a change in the direction of velocity in the given time which will give the acceleration.
Formula used:
The acceleration for the particle is the rate of change of velocity or the difference of velocity divided by the difference in time:
$\vec{a } = \dfrac{\vec{v_2} - \vec{ v_1} }{ t}$

Complete answer:
We are given the initial velocity towards eastward direction as 5 m/s. The velocity changes the direction and moves along the northwards direction. Now, we handle this problem in vector addition, the magnitude of the resultant of the difference of two vectors is given as:
$v = \sqrt{5^2 + 5^2}$
$v = 5 \sqrt{2}$ m/s.
Now, we divide this by the time interval:
$v = \dfrac{5 \sqrt{2}}{10} = \dfrac{1}{\sqrt{2}} $ m/s.
The resultant of the two vectors has to be also found with the help vector subtraction. The vectors are drawn as:

This clearly implies that the direction of the resultant is in the north east direction.

The correct answer from the given set of options is option (C).

Note:
Here to get the right approach, one has to use vector addition. Velocity and acceleration, both are vector quantities. One cannot simply handle this problem magnitude-wise. Vector addition or in our case subtraction will give the magnitude of the resultant vector. The resultant vector divided by the time period gives the acceleration. One has to clearly think about the picture that has been used to describe the question. We can imagine that the particle is going towards the east and then takes a turn so that the velocity takes the northwards direction in 10 seconds.