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Step-by-Step Solution
Step 1: Identify the Known Information
• Both cars (A and B) start from rest. Thus, initial speed $u = 0$.
• Car A has constant acceleration $a_1$, and Car B has constant acceleration $a_2$.
• Let $t_1$ be the time taken by Car A to finish the race, and $t_2$ be the time taken by Car B.
• The difference in finishing times is given as $(t_2 - t_1) = t$.
• The difference in speeds at the finish is $(v_1 - v_2) = \nu$ (the quantity we want to determine).
Step 2: Write the Expressions for Final Speeds
Since both cars start from rest and accelerate uniformly, their final speeds at the finish are:
Car A: $v_1 = a_1 \, t_1$
Car B: $v_2 = a_2 \, t_2$
Given that $v_1 - v_2 = \nu$, we have:
$a_1 \, t_1 - a_2 \, t_2 = \nu \quad \ldots (1)$
Also, from the time difference $(t_2 - t_1) = t$, we can express $t_2$ as
$t_2 = t + t_1 \,.$
Step 3: Write the Expressions for Distances
Both cars travel the same distance from start to finish. With uniform acceleration from rest, the distance covered by each car is:
Car A: $x_A = \tfrac{1}{2} a_1 \, t_1^2$
Car B: $x_B = \tfrac{1}{2} a_2 \, t_2^2$
Because $x_A = x_B$, we get:
$\tfrac{1}{2} a_1 \, t_1^2 = \tfrac{1}{2} a_2 \, t_2^2 \quad \Rightarrow \quad a_1 \, t_1^2 = a_2 \, t_2^2.$
Substitute $t_2 = t + t_1$:
$a_1 \, t_1^2 = a_2 \, (t + t_1)^2.$
Step 4: Simplify to Relate $t_1$ and $t$
Taking square roots on both sides carefully, we get:
$\sqrt{a_1}\,t_1 = \sqrt{a_2}\,(t + t_1).$
Rearranging terms gives:
$\sqrt{a_1}\,t_1 - \sqrt{a_2}\,t_1 = \sqrt{a_2}\,t \quad \Rightarrow \quad t_1 \,\bigl(\sqrt{a_1} - \sqrt{a_2}\bigr) = \sqrt{a_2}\,t.$
Thus,
$t_1 = \frac{\sqrt{a_2}\,t}{\sqrt{a_1} - \sqrt{a_2}} \quad \ldots (2)$
Step 5: Determine the Difference in Speeds $\nu$
Using equation (1):
$a_1\,t_1 - a_2\,t_2 = \nu.$
Since $t_2 = t + t_1$, rewrite it as:
$a_1\,t_1 - a_2\,(t + t_1) = \nu \quad \Rightarrow \quad a_1\,t_1 - a_2\,t_1 - a_2\,t = \nu.$
This becomes:
$(a_1 - a_2)\,t_1 - a_2\,t = \nu.$
Substitute the value of $t_1$ from equation (2):
$(a_1 - a_2) \cdot \frac{\sqrt{a_2}\,t}{\sqrt{a_1} - \sqrt{a_2}} - a_2\,t = \nu.$
Notice that $a_1 - a_2 = (\sqrt{a_1} + \sqrt{a_2})(\sqrt{a_1} - \sqrt{a_2})$. Simplifying yields:
$\bigl(\sqrt{a_1} + \sqrt{a_2}\bigr)\,\sqrt{a_2}\,t - a_2\,t = \nu.$
Which simplifies further to:
$\sqrt{a_1} \,\sqrt{a_2}\,t + a_2\,t - a_2\,t = \nu \quad \Rightarrow \quad \nu = \sqrt{a_1\,a_2}\,t.$
Step 6: State the Final Result
Therefore, the difference in speeds at the finishing point is:
$\nu = \sqrt{a_1\,a_2}\,t.$
Hence, the correct answer is $\sqrt{a_1\,a_2}\,t$.
