QJEEM19Phb5ef: JEE Main 2019 Physics - Class 11 JEEM19Phb5ef Solution: In a car race on straight road, car A takes a time t less than…

For both car initial speed ( $\mu$ ) = 0

Let the acceleration of car A and car B is $a$ ₁ and $a$ ₂ respectively.

Also let the time taken to reach the finishing point for car A is t₁ and for car B is t₂.

Let at finishing point speed of car A is $v$ ₁ and speed of car B is $v$ ₂

According to the question,

t₂ $-$ t₁ = t

and    $v$ ₁ $-$ $v$ ₂ = $v$

$\Rightarrow$    $a$ ₁t₁ $-$ $a$ ₂t₂ = $v$

$\Rightarrow$    $a$ ₁t₁ $-$ $a$ ₂(t + t₁) = $v$ . . . . . . .(1)

As, Total distance covered by both car is equal.

So,   x_A = x_B

$\Rightarrow$    ${1 \over 2}{a_1}t_1^2 = {1 \over 2}{a_2}t_2^2$

$\Rightarrow$    $a$ ₁t $_1^2$ = $a$ ₂ (t + t₁)²

$\Rightarrow$    $\sqrt {{a_1}} .{t_1}$ = $\sqrt {{a_2}}$ . (t + t₁)

$\Rightarrow$    $\sqrt {{a_1}} .{t_1} - \sqrt {{a_2}} .{t_1} = \sqrt {{a_2}} .t$

$\Rightarrow$   t₁ = ${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }}\,\,\,\,\,\,\,.....(2)$

Now put the value of t₁ in equation (2),

( $a$ ₁ $-$ $a$ ₂) t₁ $-$ $a$ ₂t = $v$

$\Rightarrow$   (a₁ $-$ a₂) . ${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }} - {a_2}t = v$

$\Rightarrow$    $\left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\sqrt {{a_2}} .t - {a_2}t = v$

$\Rightarrow$    $\sqrt {{a_1}{a_2}} .t + {a_2}.t - {a_2}t = v$

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