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Classical Mechanics Level 3

A particle moves in a straight line whose acceleration a a depends on the velocity v v according to the equation a = − α v a=-\alpha\sqrt{v} , where α \alpha is a positive constant.

At the initial moment, the velocity of the particle is v 0 {v}_{0} .

If the average speed just before it stops is v 0 β \large{\frac{{v}_{0}}{\beta}} , where β \beta is a positive integer , find β \beta .


The answer is 3.

Vaibhav Prasad by Vaibhav Prasad , 21, India

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1 solution

Vaibhav Prasad
May 21, 2016

For average speed, we need total distance x x and total time t t .

a = − α v d v d t = − α v 1 v d v = − α d t ∫ v 0 0 1 v d v = − α ∫ 0 t d t − 2 v 0 = − α t 2 v 0 = α t t = 2 v 0 α \large{a=-\alpha \sqrt { v } \\ \frac { dv }{ dt } =-\alpha \sqrt { v } \\ \frac { 1 }{ \sqrt { v } } dv=-\alpha dt\\ \int _{ { v }_{ 0 } }^{ 0 }{ \frac { 1 }{ \sqrt { v } } dv=-\alpha \int _{ 0 }^{ t }{ dt } } \\ -2\sqrt { { v }_{ 0 } } =-\alpha t\\ 2\sqrt { { v }_{ 0 } } =\alpha t\\ \boxed{t=\frac { 2\sqrt { { v }_{ 0 } } }{ \alpha }} }

Also,

a = − α v d v d t = − α v d v d t ⋅ d x d x = − α v ( d x d t ) ⋅ d v d x = − α v v ⋅ d v d x = − α v v d v = − α d x ∫ v 0 0 v d v = − α ∫ 0 x d x − 2 3 ( v 0 ) 3 2 = − α x 2 3 ( v 0 ) 3 2 = α x x = 2 3 ( v 0 ) 3 2 α \large{a=-\alpha \sqrt { v } \\ \frac { dv }{ dt } =-\alpha \sqrt { v } \\ \frac { dv }{ dt } \cdot \frac { dx }{ dx } =-\alpha \sqrt { v } \\ \left( \frac { dx }{ dt } \right) \cdot \frac { dv }{ dx } =-\alpha \sqrt { v } \\ v\cdot \frac { dv }{ dx } =-\alpha \sqrt { v } \\ \sqrt { v } dv=-\alpha dx\\ \int _{ { v }_{ 0 } }^{ 0 }{ \sqrt { v } dv=-\alpha \int _{ 0 }^{ x }{ dx } } \\ -\frac { 2 }{ 3 } { \left( { v }_{ 0 } \right) }^{ \frac { 3 }{ 2 } }=-\alpha x\\ \frac { 2 }{ 3 } { \left( { v }_{ 0 } \right) }^{ \frac { 3 }{ 2 } }=\alpha x\\ \boxed{x=\frac { \frac { 2 }{ 3 } { \left( { v }_{ 0 } \right) }^{ \frac { 3 }{ 2 } } }{ \alpha }}}

Finally, x t = v 0 3 \large {\frac{x}{t}=\frac{{v}_{0}}{3}} . Hence β = 3 \beta =\boxed{\large{3}}

8 Helpful 0 Interesting 2 Brilliant 0 Confused

Nice one...

Sparsh Sarode - 5 years ago

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