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The velocity of a particle varies with t...

The velocity of a particle varies with time as per the law `vec(V) = vec(a) + vec(b)t` where `vec(a) and vec(b)` are two constant vectors. The time at which velocity of the particle is perpendicular to velocity of the particle at t= 0 is

A

`-(|vec(a)|)/(vec(a).vec(b))`

B

`-(|vec(a)|^(2))/(vec(a).vec(b))`

C

`-(vec(a).vec(b))/(|vec(a)|^(2))`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the time \( t \) at which the velocity of the particle becomes perpendicular to its initial velocity. Let's break down the steps: ### Step 1: Understand the given equation The velocity of the particle is given by: \[ \vec{V} = \vec{a} + \vec{b} t \] where \( \vec{a} \) and \( \vec{b} \) are constant vectors. ### Step 2: Find the initial velocity At \( t = 0 \): \[ \vec{V}_0 = \vec{a} + \vec{b} \cdot 0 = \vec{a} \] Thus, the initial velocity \( \vec{V}_0 \) is simply \( \vec{a} \). ### Step 3: Express the velocity at time \( t \) At any time \( t \), the velocity is: \[ \vec{V}_t = \vec{a} + \vec{b} t \] ### Step 4: Set up the condition for perpendicularity For the velocities \( \vec{V}_0 \) and \( \vec{V}_t \) to be perpendicular, their dot product must be zero: \[ \vec{V}_0 \cdot \vec{V}_t = 0 \] Substituting the expressions we found: \[ \vec{a} \cdot (\vec{a} + \vec{b} t) = 0 \] ### Step 5: Expand the dot product Expanding the dot product gives: \[ \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} t = 0 \] Let \( |\vec{a}|^2 = \vec{a} \cdot \vec{a} \) (the magnitude squared of \( \vec{a} \)) and \( \vec{a} \cdot \vec{b} \) be the dot product of \( \vec{a} \) and \( \vec{b} \). Thus, we can rewrite the equation as: \[ |\vec{a}|^2 + (\vec{a} \cdot \vec{b}) t = 0 \] ### Step 6: Solve for \( t \) Rearranging the equation gives: \[ (\vec{a} \cdot \vec{b}) t = -|\vec{a}|^2 \] Now, solving for \( t \): \[ t = -\frac{|\vec{a}|^2}{\vec{a} \cdot \vec{b}} \] ### Conclusion The time at which the velocity of the particle is perpendicular to its initial velocity is: \[ t = -\frac{|\vec{a}|^2}{\vec{a} \cdot \vec{b}} \]
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