The acceleration of a particle is increasing linearly with time $t$ as $b t$ . The particle starts from the origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time $t$ will be

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Answer 1

here it is given that $a = b t$
wkt, $a = \frac{c h a n g e i n v e l o c i t y}{c h a n g e i n t i m e} = \frac{d v}{d t}$
$∴ \frac{d v}{d t} = b t$ now integrating this we will get $v$
$v = \frac{b t ^{2}}{2} + c$
At $t = 0 s, v = v_{0}, T h u s v_{0} = c$
$∴ v = \frac{b t ^{2}}{2} + v_{0}$
Now. $v = \frac{c h a n g e i n p o s i t i o n}{c h a n g e i n t i m e} = \frac{d s}{d t} = \frac{b t ^{2}}{2} + v_{0}$
$∴ d s = (\frac{b t ^{2}}{2} + v_{0}) d t$
$∴$ Integrating the above we get, $s = \frac{b t ^{3}}{6} + v_{0} t + c^{^{'}} a t t = 0, s = 0 ∴ c^{'} = 0$
i.e. $s = \frac{b t ^{3}}{6} + v_{0} t$

Answer 2

here it is given that

a = b t

wkt,

a = \frac{c h a n g e i n v e l o c i t y}{c h a n g e i n t i m e} = \frac{d v}{d t}

∴ \frac{d v}{d t} = b t

now integrating this we will get

v

v = \frac{b t ^{2}}{2} + c

At

t = 0 s, v = v_{0}, T h u s v_{0} = c

∴ v = \frac{b t ^{2}}{2} + v_{0}

Now.

v = \frac{c h a n g e i n p o s i t i o n}{c h a n g e i n t i m e} = \frac{d s}{d t} = \frac{b t ^{2}}{2} + v_{0}

∴ d s = (\frac{b t ^{2}}{2} + v_{0}) d t

∴

Integrating the above we get,

s = \frac{b t ^{3}}{6} + v_{0} t + c^{^{'}} a t t = 0, s = 0 ∴ c^{'} = 0

i.e.

s = \frac{b t ^{3}}{6} + v_{0} t

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