Question

If ${{v}^{2}}={{u}^{2}}+2as$, then $u=?$
A. ${{v}^{2}}-2as$
B. $\pm \sqrt{2as-{{v}^{2}}}$
C. $\pm \sqrt{{{v}^{2}}-2as}$
D. $\pm \sqrt{{{v}^{2}}+2as}$

Answer 1

Hint: Rearrange the terms given the expression, which is the equation of motion. Take the square root on both sides and find the value of initial velocity, u.

Complete step-by-step answer:
Given is the expression ${{v}^{2}}={{u}^{2}}+2as$, the equation of motion of an object where S is the distance covered by any object moving with a uniform acceleration.
We know the formula $v=u+at$, where ‘u’ is the initial velocity, ‘a’ is the acceleration of the object, ‘v’ is the final velocity and ‘t’ is the time taken by the object to cover the distance.

& v=u+at \\\ & \Rightarrow at=v-u \\\ & t=\dfrac{v-u}{a} \\\ \end{aligned}$$ So, distance travelled = (average velocity) $$\times $$time Average velocity = $$\dfrac{v+u}{2}$$= $$\dfrac{InitialVelocity+FinalVelocity}{2}$$ $\therefore S=Distance travelled \Rightarrow\left( \dfrac{v+u}{2} \right)\times \left( \dfrac{v-u}{a} \right) \\\ \Rightarrow{\dfrac{\left( v+u \right)\left( v-u \right)}{2a}} \\\ $ We know $$(a+b)(a-b)={{a}^{2}}-{{b}^{2}}.$$ $$\begin{aligned} & \therefore S=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a} \\\ & \Rightarrow {{v}^{2}}-{{u}^{2}}=2as \\\ \end{aligned}$$ Hence we got the equation of motion. Now let us rearrange them. $$\begin{aligned} & {{v}^{2}}-{{u}^{2}}=2as \\\ & {{u}^{2}}={{v}^{2}}-2as \\\ & \therefore u=\pm \sqrt{{{v}^{2}}-2as} \\\ \end{aligned}$$ [Taking root on both sides] Hence $$u=\pm \sqrt{{{v}^{2}}-2as}.$$ Option C is the correct answer. Note: There are basically 3 equations of motion. (i) $$v=u+at$$ (ii) $$s=ut+\dfrac{1}{2}a{{t}^{2}}$$ (iii) $${{v}^{2}}={{u}^{2}}+2as$$ We can use them to solve kinematic problems. We have to just identify parameters and choose appropriate equations.Hence, these equations are used to derive the components like displacement(s), velocity (initial and final), time(t) and acceleration(a). Therefore they can only be applied when acceleration is constant and motion is a straight line.