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Physics Question on Motion in a plane

A particle is moving eastward with velocity 5 m s−15 \, m \, s^{-1} In 10 s the velocity changes to 5 m s−15\, m\, s^{-1} northwards. The average acceleration in this time is

A

12ms−2\frac{1}{\sqrt{2}} m s^{-2} towards North West

B

12ms−2\frac{1}{2} m s^{-2} towards North West

C

12ms−2\frac{1}{\sqrt{2}} m s^{-2} towards North East

D

12ms−2\frac{1}{2} m s^{-2} towards North East

Answer

12ms−2\frac{1}{\sqrt{2}} m s^{-2} towards North West

Explanation

Solution

Here, v1⃗=5i^ m s−1,v2⃗=5j^ m s−1,t=10 s \vec{v_1} = 5\hat{i} \, m \, s^{-1} , \vec{v_2} = 5 \hat{j} \, m\, s^{-1}, t = 10 \, s
a⃗=v2⃗−v1⃗1=5j^−5i^10=12(j−i)\vec{a} = \frac{\vec{v_2} - \vec{v_1}}{1} = \frac{5 \hat{j} - 5 \hat{i}}{10} = \frac{1}{2} ( j - i)
∣a⃗∣=12(−1)2+(1)2| \vec{a} | = \frac{1}{2} \sqrt{( -1)^2 + (1)^2}
=122=12m s−2= \frac{1}{2} \sqrt{2} = \frac{1}{\sqrt{2}} m \, s^{-2}
tan⁡θ=ayax=(12−12)=−1\tan \theta=\frac{a_{y}}{a_{x}} = \left(\frac{\frac{1}{2}}{-\frac{1}{2}}\right)=-1
θ=tan⁡−1(−1)=135∘\theta=\tan^{-1} \left(-1\right) =135^\circ
∴   a⃗=12m s−1\therefore \:\:\: \vec{a} = \frac{1}{\sqrt{2}} m \, s^{-1} towards North West