KinematicsHard
Question
The position vector of a particle is given as = (t2 - 4 + 6)î + (t2)ĵ. The time after time which the velocity vector and acceleration vector becomes perpendicular to each is equal to : -
Options
A.1 sec
B.2 sec
C.1.5 sec
D.Not possible
Solution
= (t2 + 4t + 6) î + t2 ĵ ; v = (2t - 4) î + 2t ĵ
= 2(î + ĵ) ; when ⊥
then . = 0 ; t = 1s
Create a free account to view solution
View Solution FreeMore Kinematics Questions
From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit...The acceleration-time graph for a body is shown in the graph. Which of the following graphs would probably represent vel...A person drops from a building of height 20 m. same instant the front end of a tuck passes below the building moving wit...The velocity - time graph of the particle moving along a straight line is shown. The rate of acceleration and decelerati...A projectile is projected at an angle a (> 45o) with an initial velocity u. The time t at which its horizontal compon...