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Question: A body lying initially at point (3, 7) starts moving with a constant acceleration of 4\(\hat i\). It...

A body lying initially at point (3, 7) starts moving with a constant acceleration of 4i^\hat i. Its position after 3 s is given by the co-ordinates:
A. (7, 3)
B. (7, 18)
C. (21, 7)
D. (3, 7)

Explanation

Solution

The acceleration is only along the x – axis that is 4i^\hat i therefore the y-coordinate will remain constant then using the equation of displacement we will get the right answer.
The rate of change in the velocity of a particle with respect to time is called its acceleration. If the velocity of the particle varies at a constant rate, this phenomenon is called constant acceleration.

Complete step-by-step answer:
The initial coordinate of the particle is (3,7)
X-coordinate is unit 3
Y-coordinate unit is 7

As the body begins at rest so initial velocity=0
Body acceleration along x-axis=4units
The body's acceleration along y-axis=0
So, 3 seconds on,
Use of motion equation along the x-axis,
Sx=S0x+ut+12at2{S_x} = {S_{0x}} + ut + \dfrac{1}{2}a{t^2}
as u=0, a = 4 units, t = 3 seconds
Sx=3+12(4)32 Sx=3+18=21  {S_x} = 3 + \dfrac{1}{2}(4){3^2} \\\ {S_x} = 3 + 18 = 21 \\\
As there is no motion along y-axis so it will remain unchanged
Hence, the final coordinates are (21,7).
So, the right option is C.

Note: Whenever you face such a problem you need to consider the motion in every direction and analyse it carefully. You need to know that the equation Sx=S0x+ut+12at2{S_x} = {S_{0x}} + ut + \dfrac{1}{2}a{t^2} and other equations of motions related to this are only applicable when acceleration is constant.