The position of a particle is given by

$$r = 3.0t \;\hat{i} – 2.0t^{2} \;\hat{j} + 4.0 \hat{k} \; \;m$$

Where t is in seconds and the coefficients have the proper units for r to be in meters.

(a) Find the ‘v’ and ‘a’ of the particle?

(b) What is the magnitude and direction of the velocity of the particle at t = 2.0 s?

Asked by Abhisek | 1 year ago |  76

1 Answer

Solution :-

(a) The position of the particle is given by:

$$\vec{r} = 3.0t \;\hat{i} – 2.0t^{2} \;\hat{j} + 4.0 \hat{k}$$

Velocity $$\vec{v}$$, of the particle is given as:

$$\\\vec{v} = \dfrac{\vec{dr}}{dt} = \dfrac{d}{dt}\left ( 3.0t \;\hat{i} – 2.0t^{2} \;\hat{j} + 4.0 \hat{k} \right )\\ \\ \vec{v} = 3.0\; \hat{i} – 4.0t \; \hat{j}$$

Acceleration $$\vec{a}$$, of the particle is given as:

$$\\\vec{a} = \dfrac{d\vec{v}}{dt} = \dfrac{d}{dt}\left ( 3.0\; \hat{i} – 4.0t \; \hat{j} \right ) \\ \\ \vec{a} = -4.0 \hat{j}$$

8.54 m/s, $$69.45^{\circ}$$below the x – axis

(b) we have velocity vector,$$\vec{v} = 3.0\; \hat{i} – 4.0t \; \hat{j}$$

At t = 2.0 s:

$$\vec{v} = 3.0\; \hat{i} – 8.0 \; \hat{j}$$

The magnitude of velocity is given by:

$$\left |\vec{v} \right | = \sqrt{3.0^{2} + – 8.0^{2}} = \sqrt{73} = 8.54 \;m/s$$

Direction, $$\theta$$ = $$\tan ^{-1} \left ( \dfrac{v_{y}}{v_{x}} \right )\\ \\ = \tan ^{-1} \left ( \dfrac{-8}{3} \right ) = – \tan ^{-1} \left ( 2.667 \right )\\ \\ = 69.45^{\circ}$$

The negative sign indicates that the direction of velocity is below the x – axis.

Answered by Pragya Singh | 1 year ago

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