Question

do it with explanation​

Answer 1

Kinematics in 3-dimensions

• To solve the problem, we can consider the motion independently along all the three dimensions.

• Mass m = 3 kg
• Force $\vec{F} = 4\hat{\imath}+8\hat{\jmath}+10\hat{k}$
• Time t = 3 s
• Initial Velocity u = 0 m/s

Consider the motion individually along the x, y and z directions.

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The X-Direction

Here, $F_x = 4\ N$

So, the acceleration would be

$a_x = \dfrac{F_x}{m} = \dfrac{4}{3}\ m/s^2$

Hence, the displacement along x-direction can be calculated as:

$x = u_xt+\dfrac{1}{2}a_xt^2 \\\\\\ \implies x=0+\dfrac{1}{2}\times \dfrac{4}{3}\times 3^2 \\\\\\ \implies \boxed{x = 6\ m}$

[Note that we have already got the Answer. Option 2: (6, 12, 15). We will still solve the question for the sake of completion.]

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The Y-Direction

Here, $F_y = 8\ N$

So, the acceleration would be

$a_y = \dfrac{F_y}{m} = \dfrac{8}{3}\ m/s^2$

Hence, the displacement along y-direction can be calculated as:

$y = u_yt+\dfrac{1}{2}a_yt^2 \\\\\\ \implies y=0+\dfrac{1}{2}\times \dfrac{8}{3}\times 3^2 \\\\\\ \implies \boxed{y = 12\ m}$

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The Z-Direction

Here, $F_z = 10\ N$

So, the acceleration would be

$a_z = \dfrac{F_z}{m} = \dfrac{10}{3}\ m/s^2$

Hence, the displacement along z-direction can be calculated as:

$z = u_zt+\dfrac{1}{2}a_zt^2 \\\\\\ \implies z=0+\dfrac{1}{2}\times \dfrac{10}{3}\times 3^2 \\\\\\ \implies \boxed{x = 15\ m}$

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Thus, we have our new co-ordinates:-

x = 6 m

y = 12 m

z = 15 m

•Hence, The Answer is Option 2: (6, 12, 15)

Answer 2

Answer:

 

$\vec{F} = 4\hat{\imath}+8\hat{\jmath}+10\hat{k} \:$

$\vec{a} = \frac{4}{3} \hat{\imath}+ \frac{8}{3} \hat{\jmath}+ \frac{10}{3} \hat{k} \:$

$\vec{s} = 0+0.5 {at}^{2}$

$t = 3$

$= 4.5 \: a$

$\vec{s} = \frac{4}{3} \times4.5\hat {i}+ \frac{8}{3} \times 4.5\hat {j} + \frac{10}{3} \times 4.5\hat {k}$

$= 6\hat{\imath}+12\hat{\jmath}+15\hat{k}$

The answer is ( 6,12,15 )

hope this helps you