A body of mass $2 kg$ moves under a force of $(2^i + 3^j + 5^k) N$ . 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

Given, $\to F = (2^i + 3^j + 5^k) N$
Time, $t = 4 sec$
Let final position after $4 s$ be, $r_{f}$ .

As body start from rest, therefore position vector initially
${\to r}_{i} = (0^i + 0^j + 0^k)$ & $\to u$ (initial velocity) $= 0$

Now, from second equation of motion:
$\to s = \to u t + \frac{1}{2} \to a t^{2}$
$r_{f} - r_{i} = \frac{1}{2} \times (\frac{2^i + 3^j + 5^k}{2}) \times (4)^{2}$

$r_{f} = r_{i} + 8^i + 12^j + 20^k$

$r_{f} = 8^i + 12^j + 20^k$

So, coordinate of the final position is $(8, 12, 20)$ . Comparing it with the given coordinates $(8, b, 20)$ . The value of $b = 12$

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