Question Video: Finding the Magnitude of the Displacement Vector | Nagwa Question Video: Finding the Magnitude of the Displacement Vector | Nagwa

# Question Video: Finding the Magnitude of the Displacement Vector Mathematics • Second Year of Secondary School

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If the position vector of a particle is π« = (π‘Β² β 4π‘ β 5)π§, then the magnitude of the displacement equals zero after οΌΏ seconds. [A] 1 [B] 2 [C] 4 [D] 5

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### Video Transcript

If the position vector of a particle is π« which is equal to π‘ squared minus four π‘ minus five multiplied by the unit vector π§, then the magnitude of the displacement equals zero after what seconds. Is it (A) one, (B) two, (C) four, or (D) five?

In this question, weβre given the position vector of a particle and need to calculate when the magnitude of the displacement is equal to zero. We recall that the displacement vector π¬ of π‘ is equal to π« of π‘ minus π« of zero. We subtract the initial position from the position at time π‘. We can calculate the initial position of the particle by substituting π‘ equals zero into our expression for the position vector. This gives us zero squared minus four multiplied by zero minus five multiplied by the unit vector π§. This simplifies to give us negative five π§.

This means that the displacement vector π¬ of π‘ is equal to π‘ squared minus four π‘ minus five multiplied by π§ minus negative five multiplied by π§. Subtracting negative five is the same as adding five. This means that the displacement vector is equal to π‘ squared minus four π‘ minus five plus five multiplied by the unit vector π§. This simplifies to π‘ squared minus four π‘ multiplied by π§.

We are asked to consider the time at which the magnitude of this displacement equals zero. We know that the magnitude of any unit vector is equal to one. We, therefore, need to calculate the value of π‘ such that π‘ squared minus four π‘ is equal to zero. Factoring out π‘ from the left-hand side of our equation, we are left with π‘ multiplied by π‘ minus four. As the product of π‘ and π‘ minus four equals zero, then either π‘ equals zero or π‘ minus four equals zero. Adding four to both sides of the second equation gives us π‘ is equal to four.

We know that π‘ equals zero corresponds to the initial position, and this is not one of the options we are given. We can, therefore, conclude that the magnitude of the displacement equals zero after four seconds. The correct answer is option (C) four.

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