Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by X P ( t )=α t +β t 2 and X Q ( t )= ft - t 2. At what time, both the buses have same velocity?

Question

Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by X P ( t )=α t +β t 2 and X Q ( t )= ft - t 2. At what time, both the buses have same velocity? (A) (α- f /1+β) (B) (α+f/2(β-1)) (C) (α+ f /2(1+β)) (D) (f-α/2(1+β))

Tardigrade Admin · Accepted Answer

XP(t)=α t+β t2 XQ=f t-t2 VP(t)=α+2 β t VQ=f-2 t VP=VQ α+2 β t=f-2 t t=(f-α/2 β+2)

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

$\frac{α - f}{1 + β}$

$\frac{α + f}{2 ( β - 1 )}$

$\frac{α + f}{2 ( 1 + β )}$

$\frac{f - α}{2 ( 1 + β )}$

$X_{P} (t) = α t + β t^{2}$
$X_{Q} = f t - t^{2}$
$V_{P} (t) = α + 2 βt$
$V_{Q} = f - 2 t$
$V_{P} = V_{Q}$
$α + 2 βt = f - 2 t$
$t = \frac{f - α}{2 β + 2}$

Q. Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by XP​(t)=αt+βt2 and XQ​(t)=ft−t2. At what time, both the buses have same velocity?

1+βα−f​

2(β−1)α+f​

2(1+β)α+f​

2(1+β)f−α​