Question

A particle moves a distance x in time t according to equation x= (t+5)^-1. The acceleration of particle is proportion to?
(1) (Velocity)^3/2
(2) (Distance)²
(3) (Distance)^-2
(4) (Velocity)^2/3
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Answer 1

ans is option 1

given distance x = ( t+5)^-1

velocity v = DX / dt

= -( t+5) ^-2

acceleration a = DV / dt

= -2(t+5) ^-3

v ^3/2 = - (t+5) ^-3

acceleration. a α v^3/2

Answer 2

Given distance

x=(t+5)

−1

velcoityv=

dt

dx

=−(t+5)

−2

accelerationa=

dt

dv

=−2(t+5)

−3

v

2

3

=−(t+5)

−3

accelerationaαv

2

3

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