Two cars P and Q start from a point at the same time in a straight line and their positions are represented by XP(t)=at+bt2 and XQ(t)=ft-t2 . At what time do the cars have the same velocity?

Question

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by XP(t)=at+bt2 and XQ(t)=ft-t2 . At what time do the cars have the same velocity? (A) (a - f/1 + b) (B) (a + f/2 (b - 1)) (C) (a + f/2 (1 + b)) (D) (f - a/2 (1 + b))

Tardigrade Admin · Accepted Answer

vP=(d XP/d t)=a+2bt vQ=(d XQ/d t)=f-2t Now for, vP=vQ ⇒ a+2bt=f-2t ⇒ 2t+2bt=f-a ⇒ t=(f - a/2 (. b + 1 .))

Q. Two cars $P$ and $Q$ start from a point at the same time in a straight line and their positions are represented by $X_{P} (t) = a t + b t^{2}$ and $X_{Q} (t) = f t - t^{2}$ . At what time do the cars have the same velocity?

$\frac{a - f}{1 + b}$

$\frac{a + f}{2 ( b - 1 )}$

$\frac{a + f}{2 ( 1 + b )}$

$\frac{f - a}{2 ( 1 + b )}$

$v_{P} = \frac{d X _{P}}{d t} = a + 2 b t$
$v_{Q} = \frac{d X _{Q}}{d t} = f - 2 t$
Now for, $v_{P} = v_{Q}$
$\Rightarrow a + 2 b t = f - 2 t$
$\Rightarrow 2 t + 2 b t = f - a$
$\Rightarrow t = \frac{f - a}{2 ( b + 1 )}$

Q. Two cars P and Q start from a point at the same time in a straight line and their positions are represented by XP​(t)=at+bt2 and XQ​(t)=ft−t2 . At what time do the cars have the same velocity?

1+ba−f​

2(b−1)a+f​

2(1+b)a+f​

2(1+b)f−a​