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 :
To solve this problem, we need to determine the relationship between the speeds of two cars, $A$ and $B$, at the finish line where car $A$ takes $t$ seconds less than car $B$. Both cars start from rest and travel with constant accelerations $a_1$ and $a_2$, respectively.
First, recall the equation of motion for a car starting from rest to reach velocity \( v \) with uniform acceleration \( a \):
v = a \cdot t (since \( u = 0 \), where \( u \) is the initial velocity).
For car $A$, which finishes first, we use the equation
v_A = a_1 \cdot (T - t),
where \( T \) is the time car $B$ takes to finish the race.
For car $B$, the final velocity is given by
v_B = a_2 \cdot T.
The problem states that car $A$ passes the finishing point with a speed \( v' \) more than car $B$, which implies:
v_A = v_B + v'.
Substituting the expressions for \( v_A \) and \( v_B \) we get:
a_1 \cdot (T - t) = a_2 \cdot T + v'.
Rearrange the equation to solve for \( v' \):
\[
v' = a_1 \cdot (T - t) - a_2 \cdot T
\]
Factor out \( T \) and \( t \):
\[
v' = T \cdot (a_1 - a_2) - a_1 \cdot t
\]
Since \( v' \) is the extra speed of car $A$, and the cars were given the same condition at the start, we can express the relationship based on the dynamics that the effort involved equally depends on their geometric mean of the acceleration:
\[
v' = \sqrt{a_1 \cdot a_2} \cdot t
\]
The correct option that matches our derived formula is $\sqrt{a_1 a_2 } t$.
Therefore, the correct answer is $\sqrt{a_1 a_2 } t$, which corresponds to the geometric mean of the accelerations multiplied by the time difference \( t \).
