Question

The initial position of an object at rest is given by 3\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} - 8\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} it moves constant acceleration and reaches to the position 2\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + 4\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} after $4s$. What is its acceleration?
A. - \dfrac{1}{8}\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + \dfrac{3}{2}\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j}
B. 2\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} - \dfrac{1}{8}\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j}
C. - \dfrac{1}{2}\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + 8\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j}
D. 8\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} - \dfrac{3}{2}\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j}

Answer 1

To solve this question, we need to use the second kinematic equation of motion in the vector form. We have to substitute the initial velocity, acceleration and the displacement in the vector form. On equating the coefficients of \overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} and \overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} of both the sides of the vector equation thus obtained, we will get the required value of the acceleration in the vector form.

Formula used: The formula used to solve this question is given by
$s = ut + \dfrac{1}{2}a{t^2}$, here $u$ is the initial velocity, $s$ is the displacement, $a$ is the acceleration, and $t$ is the time.

Complete step-by-step solution:
Let the acceleration of the particle be given in the vector form by
\vec a = x\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + y\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} (1)
According to the question, at time $t = 0$, the object is at rest. So the initial velocity of the particle should be zero, that is,
$\vec u = 0$(2)
Now, the initial position of the object is given to be
${\vec r_1} = 3\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{i} - 8\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{j}$(3)
Also, the final position of the object is given to be
${\vec r_2} = 2\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{i} + 4\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{j}$(4)
So the displacement of the particle can be written as
$\vec s = {\vec r_2} - {\vec r_1}$
Putting (3) and (4) in the above equation, we get
\vec s = \left( {2\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + 4\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} } \right) - \left( {3\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} - 8\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} } \right)
\Rightarrow \vec s = - \overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + 12\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} __(5)
Also, the time taken by the particle to cover this displacement is equal to $4s$, that is,
$t = 4s$ __________(6)
Now, from the second kinematic equation, we have
$s = ut + \dfrac{1}{2}a{t^2}$
Writing this in the vector form, we have
$\vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}$
Substituting (1), (2), (5) and (6) in the above equation, we have
- \overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + 12\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} = 0 \times 4 + \dfrac{1}{2} \times \left( {x\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + y\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} } \right) \times {4^2}
\Rightarrow 8x\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + 8y\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} = - \overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + 12\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j}
Equating the coefficients of \overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} of both the sides, we get
$8x = - 1$
$\Rightarrow x = - \dfrac{1}{8}m{s^{ - 2}}$____(7)
Similarly equating the coefficients of \overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j} of both the sides, we get
$8y = 12$
$\Rightarrow y = \dfrac{3}{2}m{s^{ - 2}}$ _________(8)
Putting (7) and (8) in (1) we finally get the acceleration as
\vec a = - \dfrac{1}{8}\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{i} + \dfrac{3}{2}\overset{\lower0.5em\hbox{\smash{\scriptscriptstyle\frown}}}{j}
Hence, the correct answer is option A.

Note: Instead of using the second equation of motion in the vector form, we could also apply it separately for the two directions along x and y axis. By that way also we could get the components of the acceleration.