Question

30. A particle is projected with a velocity
v = (3i - j +2k) m/s and a constant acceleration
acting on the particle is a = (-6î +29 - 4Â) m/s.
Then the path of projectile is
7 Straight line (2) Circle
(3) Parabola
(4) Ellipse​

Answer 1

Answer:

(1) Straight line

Explanation:

Question is attached

1) We have,

$\vec{v}=(3i-j+2k)\\ \\ \vec{a}=(-6i+2j-4k)$

2) We know,

Angle between two vectors is given by :

$cos(\theta) = \frac{\vec{a}.\vec{v}}{|\vec{a}|.|\vec{v}|} \\ \\=\frac{3*-6+(-1)*2+2*-4}{\sqrt{3^2+1^2+2^2} \sqrt{6^2+2^2+4^2}} \\ \\=\frac{-28}{\sqrt{14}*\sqrt{56}}\\ \\=-1\\ \\=>\theta = \pi$

3) Hence,

Angle between acceleration and velocity vector is 180 degree.

=>  Path of projectile is straight line.

Answer 2

Answer:

straight line

Explanation:

these are parallel and opposite vectors so their some is 180° and motion is straight line and to prove them as parallel just find the ratio and that will be in whole numbers