Test: Motion in a Plane with Constant Acceleration (May 19) - JEE MCQ

 Displacement vector from A to B is given by

Let θ be angle which the direction ofmakes with the

From the previous question, 

Acceleration of the particle remains constant at all times.

The relation (b) is true, others are false because relations (a), (c) and (d) hold only for uniformly accelerated motion.

Displacement of each girl = 
Magnitude of displacement for each girl
For girl C, displacement = actual length of path.

Initial position vector of the particle,
Let final position vector of the particle be

Hence, the final coordinates of the particle are ((10, 8, 2)

The path followed by the motorist is a regular hexagon ABCDEF of side 500 m, as shown in the given figure. Let the motorist starts from A and takes the third turn at D.

Therefore, the magnitude of this displacement is 
​AD = AG + GD = 500 m + 500 m = 1000 m 

Resolving displacement 30√2 m south-west into two rectangular components, we get
Displacement in south

∴ Effective displacement due to 30 m north and 30 m south is zero.
Also effective displacement due to 30 m east and 30 m west in zero.