CBSE Class 11-science Answered
A particle is projected with a velocity v= (3i – j + 2k) m/s and a constant acceleration acting on the particle is a =( –6i+2j–4k) m/s^2. Then the path of projectile is:- (1) Straight line, (2) Circle, (3) Parabola, (4) Ellipse
Asked by patra04011965 | 11 Jan, 2020, 12:52: PM
Expert Answer
It can be seen from the ratio of vector components, velocity and acceleration is in same direction.
Since acceleration is constant. not varying with time and acting in the direction of veleocity,
particle will move in straight line
Answered by Thiyagarajan K | 11 Jan, 2020, 02:07: PM
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