हिंदी

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. -

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प्रश्न

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?

विकल्प

  • `(alpha-"f")/(1+beta)`

  • `(alpha+"f")/(2(beta-1)`

  • `(alpha+"f")/(2(1+beta)`

  • `("f"-alpha)/(2(1+beta)`

MCQ
रिक्त स्थान भरें

उत्तर

`bb(("f"-alpha)/(2(1+beta))`

Explanation:

For bus P

xp (t)  = αt + βt2

Vp (t) = α + 2βt   `[∴ "V"_"p" = ("d"x_"q")/"dt"  ]`

For bus Q

xq(t) = ft - t2

Vq (t) = f - 2t  `[∵ "V"_"q"=("d"x_"q")/"dt"]`

As, Vp (t) = Vq(t)

⇒ α + 2βt = f - 2t

⇒ f - α = 2βt + 2t

⇒ t = `("f"-alpha)/((2beta+2))` 

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