Question 149: Motion in 1D

Question

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

Accepted Answer

x = $${1 \over {t + 5}}$$
$$ 	herefore $$ v = $${{dx} \over {dt}}$$ = $${{ - 1} \over {{{\left( {t + 5} ight)}^2}}}$$ .......(i)
$$ 	herefore $$ $$a$$ = $${{dv} \over {dt}}$$ = $${2 \over {{{\left( {t + 5} ight)}^3}}}$$ = 2x
3
.......(ii)
$$ 	herefore $$ $$a$$ $$ \propto $$ (distance)
3
From equation (i), we get
$${v^{3/2}} = {{ - 1} \over {{{\left( {t + 5} ight)}^3}}}$$
Putting this in equation (ii), we get
$$a$$ = -2$${v^{3/2}}$$
$$ 	herefore $$ $$a$$ $$ \propto $$ (velocity)
3/2

Answer

(velocity)
3/2

Answer

(distance)
2

Answer

(distance)
$$-$$2

Answer

(velocity)
2/3

Motion in 1D

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

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