Question

The velocity of a particle varies with time as por law v=a+bt where i and 6 are two constant
vectors. The time at which velocity of the particle is perpendicular to the velocity of the perides
t = 0 is
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laf
á b
(B)
(D) None of these
a.b
ab
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Answer 1

Answer:

Ans is option D

Explanation:

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Answer 2

Answer:

The time at which the velocity is perpendicular to velocity at t=0 is given as

$t = - \frac{ |a| {}^{2} }{ |a.b| }$

Explanation:

Given that,

The velocity of particle varies as v=a+bt,

where a and b are vectors

Velocity of particle at t=0 is given as

$v_{0} = a + b(0) = a$

Velocity of particle at time t=t is given as

$v_{t} = a + bt$

According to given question,both these velocities are perpendicular hence their dot products must be zero.

Therefore

$v_{0}. v_{t} = 0 \\ a.(a + bt) = 0 \\ a.a + (a.b)t = 0$

Now equating the magnitudes of the dot products we obtain an relation for time t as

$t = - \frac{ |a| {}^{2} }{ |a.b| }$

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