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Question: The relation between time and distance is \(t = \alpha x ^ { 2 } + \beta x\) , where \(\alpha\) a...

The relation between time and distance is t=αx2+βxt = \alpha x ^ { 2 } + \beta x , where α\alpha and β\beta are constants. The retardation is

A

2αv32 \alpha v ^ { 3 }

B

2βv32 \beta v ^ { 3 }

C

2αβv32 \alpha \beta v ^ { 3 }

D

2β2v32 \beta ^ { 2 } v ^ { 3 }

Answer

2αv32 \alpha v ^ { 3 }

Explanation

Solution

dtdx=2αx+β⇒v=12αx+β\frac { d t } { d x } = 2 \alpha x + \beta \Rightarrow v = \frac { 1 } { 2 \alpha x + \beta }

∙∙\bullet \bullet a=dvdt=dvdx⋅dxdta = \frac { d v } { d t } = \frac { d v } { d x } \cdot \frac { d x } { d t }

a=vdvdx=−v⋅2α(2αx+β)2=−2α⋅v⋅v2=−2αv3a = v \frac { d v } { d x } = \frac { - v \cdot 2 \alpha } { ( 2 \alpha x + \beta ) ^ { 2 } } = - 2 \alpha \cdot v \cdot v ^ { 2 } = - 2 \alpha v ^ { 3 }

∴\therefore Retardation =2αv3= 2 \alpha v ^ { 3 }