Question

The initial position of an object at rest is given by $3\widehat{i}-8\widehat{j}$. It moves with constant acceleration and reaches to the position \[2\widehat{i}+4\widehat{j}\] after \[4s\]. What will be its acceleration?

Answer 1

Hint: First of all the initial and the final positions on the X and Y axis is to be found. The difference between the initial and final positions on each axis will be the product of the half of the acceleration in that axis and the square of the time. This can be found on both the axes. From this we will get the values of accelerations. By writing these values in the form of a vector expression will give the answer hope these may help you to solve this question.

Complete step by step answer:
The initial position of the object on the axis is given by the form,
\[I=3\widehat{i}-8\widehat{j}\]
From this, the initial position of the object on the x-axis is can be found, given as,
\[{{x}_{i}}=3\]
And the initial position on the y-axis is given as,
\[{{y}_{i}}=-8\]
Now the second polar form is given for indicating the final position of the object is given as,
\[F=2\widehat{i}+4\widehat{j}\]
From this, the final position of the object on the x-axis is given by the equation,
\[{{x}_{f}}=2\]
And the final position of the object on the y-axis is given by the equation,
\[{{y}_{f}}=4\]
As we all know, the change in position is given by the equation,
\[S=\dfrac{1}{2}a{{t}^{2}}\]
Where \[S\]be the change in position, \[a\]be the acceleration of the object and \[t\]be the time taken. Using this relation on the x- axis we can write that,
\[{{x}_{f}}-{{x}_{i}}=\dfrac{1}{2}{{a}_{x}}{{t}^{2}}\]
The time has been given as,
\[t=4s\]
Substituting the values in it will give,
\[2-3=\dfrac{1}{2}{{a}_{x}}{{\left( 4 \right)}^{2}}\]
Rearranging this equation will give the acceleration in x-axis,
\[{{a}_{x}}=\dfrac{-2}{16}=\dfrac{-1}{8}\]
Similarly in the y-axis, the relation can be expressed as,
\[{{y}_{f}}-{{y}_{i}}=\dfrac{1}{2}{{a}_{y}}{{t}^{2}}\]
Rearranging and substituting the values in it will give,
\[{{a}_{y}}=\dfrac{24}{16}=\dfrac{3}{2}\]
Therefore the acceleration can be written as,
\[\vec{a}=\dfrac{-1}{8}\widehat{i}+\dfrac{3}{2}\widehat{j}\]

Note: Acceleration is given as the rate of change of velocity with respect to the time interval taken. This is a vector quantity in which the magnitude of the acceleration in y axis is given by the sine component and the x-component is obtained using the cosine part.