Three forces ${\to F}_{1} = (2^i + 4^j) N; {\to F}_{2} = (2^i -^4 j) N$ and $v e c F_{3} = (^k - 4^i - 2^j) N$ are applied on an object of mass 1 kg at rest at origin. The position of the object at $t = 2 s$ will be

4 m

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8 m

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3 m

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1 m

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Solution

Step 1, Given data
mass of the object (m) = 1 kg
Initially, object is at rest so u = 0
Step 2 Finding the acceleration of the object
Net force is the sum of all the forces acting on the object
$\to F = (2 + 2 - 4) \to i + (4 - 4 - 2) \to j (0 + 0 + 1) \to k = 0 \to i - 2 \to j + \to k$
We know,
$\to F = m \to a$
So acceleration will be $\to a = \frac{\to F}{m} = \frac{0 \to i + 2 \to j + \to k}{1 k g} = (0 \to i + 2 \to j + \to k)$ m/ $s^{2}$
Step 3, Finding the position
Position of the body in x-direction after 2 second
We know,
$S = u t + \frac{1}{2} a t^{2}$
initial velocity is zero
So,
$x = \frac{1}{2} \times 0 \times 2^{2} = 0$
Position of the body in y-direction after 2 seconds
$y = \frac{1}{2} \times - 2 \times 4 = - 4$
Position of the body in z direction
$z = \frac{1}{2} \times 1 \times 4 = 2$
So, the position of the body after 2 seconds is $(0, - 4, 2)$
Now,
Total distance traveled by the object will be
$\sqrt{0^{2} + {- 4}^{2} + 2^{2}} = \sqrt{20} = 2 \sqrt{5} m$
Hence the correct option is (C)

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Three forces →F1=(2^i+4^j)N;→F2=(2^i−^4j)N and vecF3=(^k−4^i−2^j)N are applied on an object of mass 1 kg at rest at origin. The position of the object at t=2s will be

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