Single Choice

A particle moving with an initial velocity $$ \hat { i } -4\hat { j } +10\hat { k } $$ has acceleration $$ \hat { i } +\hat { j } -2\hat { k } $$. Its velocity at the end of 2 seconds, points along the unit vector:

A$$ \dfrac { 1 }{ 7 } \left( -3\hat { i } +2\hat { j } -6\hat { k } \right) $$
B$$ \dfrac { 1 }{ \sqrt { 382 } } \left( 3\hat { i } -7\hat { j } +18\hat { k } \right) $$
C$$ \dfrac { 1 }{ 7 } \left( 3\hat { i } -2\hat { j } +6\hat { k } \right) $$
Correct Answer
D$$ \dfrac { 1 }{ \sqrt { 77 } } \left( 2\hat { i } -3\hat { j } +8\hat { k } \right) $$

Solution

As we know
$$v=u+at$$
$$v=(\hat i-4\hat j+10\hat k)+(\hat i+\hat j-2\hat k)\times 2$$
$$v=(\hat i-4\hat j+10\hat k)+(2\hat i+2\hat j-4\hat k)$$
$$v=3\hat i-2\hat j+6\hat k$$
Unit vector is given as,
$$|v|=\sqrt{3^2+(-2)^2+6^2}$$
$$|v|=\sqrt{49}$$
$$|v|=7$$
So, Velocity at the end of 2 sec, points along unit vector$$=\dfrac{v}{|v|}$$
$$=\dfrac{3\hat i-2\hat j+6\hat k}{7}$$
$$=\dfrac{1}{7}(3\hat i-2\hat j+6\hat k)$$

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