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# Rate of change of speed of a particle in motion is nothing but the component of acceleration along the velocity $v$. Velocity and acceleration of particle at some instant of time are $\vec v = \left( {2\hat i + \hat j + 2\hat k} \right)\, {\text{m/s}}$ and acceleration $- \left( {\hat i + 2\hat j + \hat k} \right)\, {\text{m/}}{{\text{s}}^{\text{2}}}$. Then the speed of the particle is _______ at a rate of ________ ${\text{m/}}{{\text{s}}^{\text{2}}}$.A) Increasing, $2$B) Decreasing, $2$C) Increasing, $4$D) Decreasing, $4$

Last updated date: 16th Jul 2024
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Hint: In this question, we will use the concept of vectors that is the magnitude and the direction of the vector form of the acceleration. First, we will discuss the basic concept of velocity and acceleration. Then calculate the value of acceleration and show its direction.

Complete step by step solution:
When a particle is moving in a direction then the rate of change in displacement is known as velocity. It is a vector quantity. Velocity is a function of time. Speed gives us the idea how fast or slow an object is moving. Velocity gives us the speed and its direction also. Acceleration is the rate of range of velocity. It is also a vector quantity.

We know that if the value of acceleration is positive the speed of a particle will increase. Whereas if the acceleration is negative then the speed of the particle will decrease.
Here, we have given the value of acceleration as $- \left( {\hat i + 2\hat j + \hat k} \right)\,{\text{m/}}{{\text{s}}^{\text{2}}}$.
It is negative in sign.
Now we will calculate the magnitude of acceleration as $\Rightarrow \sqrt {{1^2} + {2^2} + {1^2}} = \sqrt 6 {\text{m/}}{{\text{s}}^{\text{2}}}$
So, from the above calculation the acceleration of the particle is negative and its of value $\sqrt 6 \,{\text{m/}}{{\text{s}}^{\text{2}}}$.
The speed of the particle is decreasing at a rate of $\sqrt 6 \,{\text{m/}}{{\text{s}}^{\text{2}}}$.

Note: As we know that the velocity and the acceleration are the vector quantities, that is they have both magnitude and the direction. Here the velocity of the particle is in a positive direction. So the speed is decreasing. But if the velocity is given in the opposite direction then the speed of the particle would have been increasing.