A body of mass $2 k g$ moves under a force of $(2 \hat{i} + 3 \hat{j} + 5 \hat{k}) N$ . It starts from rest and was at the origin initially. After $4 s$ , its new coordinates are $(8, b, 20)$ . The value of $b$ is _______. (Round off to the Nearest Integer)

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Solution

Step 1. Given data
Mass of the body, $m = 2 k g$
Force, $F = (2 \hat{i} + 3 \hat{j} + 5 \hat{k}) N$
Initial velocity, $u = 0$
The body starts from origin.
Coordinates of final position $= (8, b, 20)$ [ $e q (1)$ ]
Step 2. Finding the value of $b$
As the body starts from rest, then its position vector initially, $r_{i} = (0 \hat{i} + 0 \hat{j} + 0 \hat{k})$
Let the position vector finally, $r_{f} = (x \hat{i} + y \hat{j} + z \hat{k})$
Using the second equation of motion,
$S = u t + \frac{1}{2} a t^{2}$ [ $S$ is the distance and $a$ is the acceleration.]
$r_{f} - r_{i} = u t + \frac{1}{2} (\frac{F}{m}) t^{2}$ [ $F = m a$ ]
$(x \hat{i} + y \hat{j} + z \hat{k}) - (0 \hat{i} + 0 \hat{j} + 0 \hat{k}) = \frac{1}{2} (\frac{(2 \hat{i} + 3 \hat{j} + 5 \hat{k})}{2}) {(2)}^{2}$
$(x \hat{i} + y \hat{j} + z \hat{k}) = 8 \hat{i} + 12 \hat{j} + 20 \hat{k}$
Comparing it with $e q (1)$ , we get
Therefore, $b = 12$ .

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