Question

For a particle moving along a straight line and its position X is given by ${X = 2 {e}^{ - 2t}}$ the way it is t is time . Relation between velocity v and acceleration a is given by
1. a = - v
2.a = v
3.a = - 2v
4.a = 2v

Only correct answers accepted ​

Answer 1

Answer:

(3) a = -2v

Explanation:

It is being given that, position of the particle in straight line varies by Following equation :-

${X = 2 {e}^{ - 2t}}$

We know that, On differentiation of position wrt time, we get velocity as a function of time.

So , differentiate the Following equation :-

We get ,

$\frac{dx}{dt} = 2 \frac{d( {e}^{ - 2t})}{dt} = 2 \times {e}^{ - 2t} \times - 2 = - 4 {e}^{ - 2t}$

Now, dx/dt = velocity

$= > v = - 4 {e}^{ - 2t}$

..................(i)

Now, again differentiate the velocity equation wrt time,

we get,

$= > \frac{dv}{dt} = - 4 \times {e}^{ - 2t} \times - 2$

So, acceleration is

$= > a = 8 {e}^{ - 2t}$

...................(ii)

on comparing equation (i) and (ii), we get

$= > a = - 2v$

Hence, option (3) is correct.

Answer 2

$\Huge{\underline{\underline{\mathfrak{Question \ \colon }}}}$

For a particle moving along a straight line and its position X is given by ${X = 2 {e}^{ - 2t}}$ the way it is t is time . Relation between velocity v and acceleration a is given by

1) a = - v

2) a = v

3)a = - 2v

4)a = 2v

$\Huge{\underline{\underline{\mathfrak{Answer \ \colon }}}}$

From the Question,

The position vector of the particle is defined by the relation :

$\large{ \tt{x = 2 {e}^{ - 2t} }}$

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To find

Relation between velocity and acceleration of the particle

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Differentiating x w.r.t t,we get velocity of the particle :

$\sf{v = \frac{dx}{dt} } \\ \\ \leadsto \: \sf{v = \frac{d(2 {e}^{ - 2t)} }{dt} } \\ \\ \huge{ \leadsto \: \boxed{ \boxed{ \tt{ \green{v = - 4 {e}^{ - 2t} \: m {s}^{ - 1} }}}}}$

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Differentiating v w.r.t to t,we get acceleration of the particle :

$\sf{a = \frac{dv}{dt} } \\ \\ \leadsto \: \sf{a = \frac{d( - 4 {e}^{ - 2t}) }{dt} } \\ \\ \huge{ \leadsto \: \boxed{ \boxed{ \tt{ \green{a = 8 {e}^{ - 2t} \: m {s}^{ - 2} }}}}}$

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Dividing values of acceleration and velocity,we get :

$\sf{ \frac{a}{v} = - \frac{8 \cancel{ {e}^{ - 2t}} }{4 \cancel{ {e}^{ - 2t} }} } \\ \\ \huge{ \leadsto \: \sf{a = - 2v}}$

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Third Option is correct

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