Question

A particle moves a distance x in time t according to
equation x = (t + 5)-¹. The acceleration of particle is
proportional to

(a) (velocity)^3/2

(b) (distance)²


(c) (distance)-²

(d) (velocity)⅔​

Answer 1

Question:-

• A particle moves a distance x in time t according to equation x = (t + 5)-¹. The acceleration of particle is proportional to

To Find:-

• Find the acceleration of the particle is proportional to .

Solution:-

• $\sf \: v = \dfrac { dx } { dt } = { -( t + 5 ) }^{ -2 }$

• $\sf \: v = \dfrac { dv } { dt } = { -2( t + 5 ) }^{ -3 }$

Then ,

• $\sf \: { v }^{ \frac { 3 } { 2 } } = \dfrac { dx } { dt } = { -( t + 5 ) }^{ -3 }$

• $\sf \: a \: \: \alpha \: \: { v }^{ \frac { 3 } { 2 } }$

Answer 2

${\pmb{\underline{\sf{Required \: Solution....}}}}$

Distance (x) = $(t + 5) {}^{ - 1}$

Velocity = $\frac{dx}{dt}$

$: \implies - 1(t + 5) {}^{ - 1 - 1} \\ \boxed{: \implies - 1(t + 5) {}^{ - 2} }$

Acceleration = $\frac{dx}{dt}$

$: \implies - 2(t + 5) {}^{ - 3} \\ \: \boxed{ : \implies - 1(t + 5) {}^{ - 3} }$

______________________________

$V {}^{ \frac{3}{2} } = - (t + 5) {}^{ - 3}$

$a∝ - (t + 5) {}^{ - 2 \times \frac{3}{2} } \\ \boxed {a∝v {}^{ \frac{3}{2} } }$

Option (a) : (Velocity)^3/2 is Correct

$\sf\blue{hope \: this \: helps \: you!! \: }$

@Mahor2111

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