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:

Let the total distance of the race be $$S$$. Both cars start from rest, so their initial velocities are zero.

For motion with uniform (constant) acceleration, the basic kinematic relations we shall use are

1. Distance covered from rest in time $$T$$: $$S=\dfrac12\,a\,T^{2}.$$

2. Final speed attained from rest in time $$T$$: $$v_{\text{final}}=a\,T.$$

Denote

$$T_B=$$ time taken by car $$B,$$

$$T_A=$$ time taken by car $$A.$$

The statement “car A takes a time $$t$$ less than car B” gives us

$$T_A=T_B-t. \quad -(1)$$

Since both cars run over the same total distance $$S$$, we write the distance formula for each:

For car $$A$$ (acceleration $$a_1$$):

$$S=\dfrac12\,a_1\,T_A^{2}.$$

For car $$B$$ (acceleration $$a_2$$):

$$S=\dfrac12\,a_2\,T_B^{2}.$$

Because the right-hand sides are both equal to the same $$S$$, we equate them:

$$\dfrac12\,a_1\,T_A^{2}=\dfrac12\,a_2\,T_B^{2}.$$

The factor $$\dfrac12$$ appears on both sides, so we cancel it, obtaining

$$a_1\,T_A^{2}=a_2\,T_B^{2}. \quad -(2)$$

Now substitute the relation (1), $$T_A=T_B-t,$$ into equation (2):

$$a_1\,(T_B-t)^{2}=a_2\,T_B^{2}.$$

We expand the square carefully:

$$a_1\left(T_B^{2}-2\,T_B\,t+t^{2}\right)=a_2\,T_B^{2}.$$

Multiplying term by term gives

$$a_1\,T_B^{2}-2\,a_1\,T_B\,t+a_1\,t^{2}=a_2\,T_B^{2}.$$

Next we bring every term to the left side to form a quadratic equation in $$T_B$$:

$$(a_1-a_2)\,T_B^{2}-2\,a_1\,T_B\,t+a_1\,t^{2}=0. \quad -(3)$$

Equation (3) is a quadratic of the standard form $$A\,T_B^{2}+B\,T_B+C=0$$ with

$$A=a_1-a_2,\quad B=-2\,a_1\,t,\quad C=a_1\,t^{2}.$$

We need $$T_B$$, so we apply the quadratic-formula statement

$$T_B=\dfrac{-B\pm\sqrt{B^{2}-4AC}}{2A}.$$

Substituting $$A,B,C$$ from above:

$$T_B=\dfrac{-(-2\,a_1\,t)\pm\sqrt{(-2\,a_1\,t)^{2}-4\,(a_1-a_2)\,(a_1\,t^{2})}}{2\,(a_1-a_2)}.$$

Simplifying step by step:

• The first numerator term: $$-(-2\,a_1\,t)=2\,a_1\,t.$$

• Inside the square root:

$$(-2\,a_1\,t)^{2}=4\,a_1^{2}\,t^{2},$$

$$4AC=4\,(a_1-a_2)\,(a_1\,t^{2})=4\,a_1\,(a_1-a_2)\,t^{2}.$$

Therefore

$$B^{2}-4AC=4\,a_1^{2}\,t^{2}-4\,a_1\,(a_1-a_2)\,t^{2} =4\,t^{2}\left(a_1^{2}-a_1(a_1-a_2)\right).$$

Inside the brackets we simplify the expression:

$$a_1^{2}-a_1(a_1-a_2)=a_1^{2}-a_1^{2}+a_1\,a_2=a_1\,a_2.$$

So the discriminant becomes

$$B^{2}-4AC=4\,t^{2}\,a_1\,a_2.$$

The square root therefore is

$$\sqrt{B^{2}-4AC}=2\,t\,\sqrt{a_1\,a_2}.$$

Putting everything back into the expression for $$T_B$$:

$$T_B=\dfrac{2\,a_1\,t\;\pm\;2\,t\,\sqrt{a_1\,a_2}}{2\,(a_1-a_2)} =\dfrac{a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}}{a_1-a_2}.$$

Only the “plus” or “minus” that keeps $$T_B$$ positive is physically acceptable; we shall keep both for the moment and the correct sign will reveal itself automatically when we calculate the speed difference.

Now we focus on the required quantity, namely the speed difference at the finish. Using the second kinematic relation $$v_{\text{final}}=a\,T$$, the individual finish speeds are

For car $$A$$: $$v_A=a_1\,T_A=a_1\,(T_B-t).$$

For car $$B$$: $$v_B=a_2\,T_B.$$

The problem defines $$v$$ as “the speed of car A at the finish is $$v$$ more than that of car B,” which translates to

$$v=v_A-v_B.$$

Substituting the expressions of $$v_A$$ and $$v_B$$, we have

$$v=a_1\,(T_B-t)-a_2\,T_B =a_1\,T_B-a_1\,t-a_2\,T_B =(a_1-a_2)\,T_B-a_1\,t. \quad -(4)$$

Equation (4) already contains the factor $$(a_1-a_2)\,T_B,$$ which suggests inserting the value of $$T_B$$ that we just derived. So we substitute $$T_B=\dfrac{a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}}{a_1-a_2}$$ into (4):

$$v=(a_1-a_2)\left(\dfrac{a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}}{a_1-a_2}\right)-a_1\,t.$$

The factor $$(a_1-a_2)$$ cancels immediately, leaving

$$v=a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}-a_1\,t=t\,\sqrt{a_1\,a_2},$$

because the $$a_1\,t$$ terms subtract out. The sign that survives is the positive one; speed difference must be positive. Thus we finally obtain

$$v=t\,\sqrt{a_1\,a_2}.$$

Looking at the given options, this matches exactly with Option C: $$\sqrt{a_1\,a_2}\;t.$$

Hence, the correct answer is Option C.