Question

How do you make $''a''$ the subject in $s=ut+\dfrac{1}{2}a{{t}^{2}}$?

Answer 1

Hint: In the formula of “s” given in the above problem i.e. $s=ut+\dfrac{1}{2}a{{t}^{2}}$. We are asked to find the expression of $''a''$ in terms of $''s,u,t''$ so we are going to rearrange the given expression in such a way so that “a” in one side of the equation and $''s,u,t''$ on the other side of the equation.

Complete step by step answer:
The expression given in the above problem that we have to rearrange is as follows:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
In the above problem, we are asked to make $''a''$ as a subject meaning $''a''$ will be on one side of the equation and $''s,u,t''$ on the other side of the equation. To achieve that we are going to subtract $ut$ on both the sides and we get,
$s-ut=\dfrac{1}{2}a{{t}^{2}}$
Now, multiplying 2 on both the sides we get,
$2\left( s-ut \right)=a{{t}^{2}}$
In the above equation, if we divide ${{t}^{2}}$ on both the sides then we will get “a” on one side of the equation and $''s,u,t''$ on the other side so dividing ${{t}^{2}}$ on both the sides we get,
$\dfrac{2\left( s-ut \right)}{{{t}^{2}}}=a$
Now, interchanging L.H.S and R.H.S of the above equation we get,
$a=\dfrac{2\left( s-ut \right)}{{{t}^{2}}}$
From the above, we have made $''a''$ as a subject in the given equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$ and the subject $''a''$ in terms of $''s,u,t''$ is equal to:

$a=\dfrac{2\left( s-ut \right)}{{{t}^{2}}}$

Note:
 You can check whether the expression that we got i.e. $a=\dfrac{2\left( s-ut \right)}{{{t}^{2}}}$ is correct or not by substituting this expression of $''a''$ in the given equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$ we get,
$s=ut+\dfrac{1}{2}\dfrac{2\left( s-ut \right)}{{{t}^{2}}}{{t}^{2}}$
In the above equation, ${{t}^{2}}$ will be cancelled from the numerator and the denominator and we get,
$s=ut+\dfrac{1}{2}\left( 2 \right)\left( s-ut \right)$
In the R.H.S of the above equation, 2 will be cancelled out from the numerator and the denominator and we get,
$s=ut+s-ut$
In the R.H.S of the above equation, $ut$ will be cancelled out and we get,
$s=s$
As you can see that L.H.S is equal to R.H.S in the above equation so the expression of $''a''$ that we got above is correct.