Question

In a car race on straight road, car $A$ takes a time $t$ less than car $B$ at the finish and passes finishing point with a speed $'v'$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $'v'$ is equal to :

• A. $\frac{a_{1} + a_{2}}{2} t $
• B. $\sqrt{2 a_1 a_2 } t$
• C. $\frac{2 a_1 a_2}{a_1 + a_2} t $
• D. $\sqrt{a_1 a_2 } t$

Answer 1

The Correct Option is D

For $A$ & $B$ let time taken by $A$ is $t_0$
from ques.
$v_{A} - v_{B} = v =\left(a_{1} -a_{2}\right)t_{0} -a_{2}t$ ....(i)
$ x_{B} =x_{A} = \frac{1}{2} a_{1} t_{0}^{2} = \frac{1}{2} a_{2} \left(t_{0} +t\right)^{2} $
$ \Rightarrow \sqrt{a_{1}t_{0}} = \sqrt{a_{2} } \left(t_{0} +t\right) $
$ \Rightarrow \left(\sqrt{a_{2}} - \sqrt{a_{2}}\right)t_{0} = \sqrt{a_{2}t} $ ......(ii)
putting $t_0$ in equation
$ v = \left(a_{1} -a_{2}\right) \frac{\sqrt{a_{2}t}}{\sqrt{a_{1} } -\sqrt{a_{2}}} -a_{2} t $
$= \left(\sqrt{a_{1}} + \sqrt{a_{2}}\right) \sqrt{a_{2}t} - a_{2} t \Rightarrow v = \sqrt{a_{1}a_{2}t} $
$ \Rightarrow \sqrt{a_{1}a_{2}t} + a_{2}t - a_{2}t $

Answer 2

An equation of motion is a mathematical formula that describes how a physical system behaves over time. 

• In terms of dynamic variables, the equation of motion specifies the motion of objects and systems. 
• Many important motion properties, including velocity, displacement, speed, time, and acceleration, are linked together by these equations.

Equations of Motion

There are three equations of motion that are only applicable to uniformly accelerated motion. The following are the equations:

1. v = u + at
2. v² = u² + 2as
3. s = ut + 1/2 at²

Where

• v is the final Velocity 
• u is the initial velocity 
• t is the time taken 
• a is constant acceleration or uniform acceleration
• s is the displacement of the body