Question

When $t=0$ a particle at (1,0,0) moves towards (4,4,12) with a constant speed of 65m/s. The position
of the particle is measured in metres and time in sec. Assuming constant velocity, the position of the
particle at $t=2$ sec is :

• A. $(13\hat{i}-120\hat{j}+40\hat{k})m$
• B. $(40\hat{i}+31\hat{j}-120\hat{k})m$
• C. $(13\hat{i}-40\hat{j}+12\hat{k})m$
• D. $(31\hat{i}+40\hat{j}+120\hat{k})m$

Answer 1

The correct option is D $(31\hat{i}+40\hat{j}+120\hat{k})m$

Given: Speed of the object (u) = 65 m /s

Position vector of initial point (x), s₁ = 1 i

Position vector of final point (Y), s₂ = 4 i+4 j+12 k

Net displacement XY = (s₂ -s₁) = (4 - 1) i+(4-0) j +(12 - 0) k = 3 i+4 j+12 k

Magnitude of XY vector, |XY| =

Let XY vector make angles α, β, and γ with the x-axis, y-axis, and z-axis respectively.

Now, we shall find their direction cosines.

cosα = XY . i / 13 |i| = 3 / 13 ---------As ( i.i =1, i.j =0 and i.k = 0) and |i| = 1

cosβ = XY . j / 13 |j| = 4 / 13 ----------As ( j.j =1, j.i = 0 and j.k = 0) and |j| = 1

cosγ = XY . k / 13 |k| = 12/ 13 ---------As ( k.k =1, k.j =0 and k.i = 0) and |k| =1

Let (x,y,z) the position coordinate of the object after 2 s, then

x = 1 + (u cos α) × 2 = 1 + [65× (3/13)×2] = 31

y = 0 + (u cos β) × 2 = 0 + [65× (4/13)×2] = 40

z = 0 + (u cos γ) × 2 = 0 + [65× (12/13)×2] = 120

Hence, the required position coordinate of the object after 2 s will be (x,y,z) =(31, 40, 120)