A particle moving with an initial velocity - -4- -10k- has acceleration - - -2k- Its velocity at the end of 2 seconds- points along the unit vector-

The correct option is C 1 7 ( 3î 2ĵ +6k̂) As we know #160; v=u+at v=(î 4ĵ+10k̂)+(î+ĵ 2k̂)× 2 v=(î 4ĵ+10k̂)+(2î+2ĵ 4k̂) v=3î 2ĵ+6k̂ ...

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Byju's Answer

Standard XII

Physics

Instantaneous Velocity

A particle mo...

Question

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

\frac{1}{7} (- 3^i + 2^j - 6^k)

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\frac{1}{\sqrt{382}} (3^i - 7^j + 18^k)

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\frac{1}{7} (3^i - 2^j + 6^k)

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\frac{1}{\sqrt{77}} (2^i - 3^j + 8^k)

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Solution

The correct option is C $\frac{1}{7} (3^i - 2^j + 6^k)$
As we know
$v = u + a t$
$v = (^i - 4^j + 10^k) + (^i +^j - 2^k) \times 2$
$v = (^i - 4^j + 10^k) + (2^i + 2^j - 4^k)$
$v = 3^i - 2^j + 6^k$
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 $= \frac{v}{| v |}$
$= \frac{3^i - 2^j + 6^k}{7}$
$= \frac{1}{7} (3^i - 2^j + 6^k)$

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Similar questions

Let $\to b = -^i + 4^j + 6^k a n d \to c = 2^i - 7^j - 10^k$ . If $\to a$ is a unit vector and the scalar triple product $[\to a \to b \to c]$ has the greatest value, then $\to a$ is equal to

The unit vector parallel to the resultant of the vectors
$\to A = 4^i + 3^j + 6^k$ and
$\to B = -^i + 3^j - 8^k$ is

Q. The unit vector which is orthogonal to the vector

3^i + 2^j + 6^k

is coplanar with vectors

2^i +^j + 6^k a n d^i -^j -^k

Q. The vector

\to a = 1 / 7 (2^i + 3^j + 6^k)

\to b = 1 / 7 (3^i - 6^j + 2^k)

\to c = 1 / 7 (6^i + 2^j - 3^k)

form

Q. The unit vector which is orthogonal to the vector

3^i + 2^j + 6^k

is coplanar with vectors

2^i +^j + 6^k a n d^i -^j -^k

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A particle moving with an initial velocity ^i−4^j+10^k has acceleration ^i+^j−2^k. Its velocity at the end of 2 seconds, points along the unit vector:

The correct option is C 17(3^i−2^j+6^k)As we know v=u+atv=(^i−4^j+10^k)+(^i+^j−2^k)×2v=(^i−4^j+10^k)+(2^i+2^j−4^k)v=3^i−2^j+6^kUnit vector is given as,|v|=√32+(−2)2+62|v|=√49|v|=7So, Velocity at the end of 2 sec, points along unit vector=v|v| =3^i−2^j+6^k7 =17(3^i−2^j+6^k)

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