Question

at t= 0, a particle at (1,0,0) moves with velocity vector = (15 i cap 20 j cap + 60 k cap)m/s . find its position vector at time = 2 sec​

Answer 1

Answer:

Explanation:

Given: Speed of the object (u) = 65 cm /s = 0.65 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 + [0.65× (3/13)×2] = 1.3

y = 0 + (u cos β) × 2 = 0 + [0.65× (4/13)×2] = 0.4

z = 0 + (u cos γ) × 2 = 0 + [0.65× (12/13)×2] = 1.2

Hence, the required position coordinate of the object after 2 s will be (x,y,z) =(1.3, 0.4, 1.2)