Question

A particle is moving in a straight line under acceleration a = kt, where k is a constant. Find the velocity in terms of t. The motion starts from rest.​

Answer 1

SoluTion:

Given acceleration:

• $\tt{a(t) = kt}$

We know that,

$\large{\boxed{\tt{\blue{a(t) = \dfrac{d}{dt} [v(t)]}}}}$

Using given expression,

$\longrightarrow$ $\tt{\dfrac{d}{dt} [v(t)] = kt}$

$\longrightarrow$ $\tt{d[v(t)] = kt.dt}$

Integrating it,

$\longrightarrow$ $\tt{v(t)}$ = $\displaystyle\int kt.dt$

$\longrightarrow$ $\tt{v(t)}$ = $k \displaystyle\int t.dt$

$\longrightarrow$ $\tt{v(t)}$ = $\tt{k. \dfrac{t^{2}}{2} + c}$

$\longrightarrow$ $\tt{v(t)}$ = $\tt{\dfrac{1}{2} kt^{2} + c}$

Also, it is given that motion starts from rest, so $\bold{\tt{v = 0}}$ at $\bold{\tt{t = 0}}$.

$\longrightarrow$ $\tt{v(0)=0}$

$\longrightarrow$ $\tt{\dfrac{1}{2}k(0)^{2}\:+c=0}$

$\longrightarrow$ $\tt{c = 0}$

Hence,

$\large{\boxed{\boxed{\pink{\sf{v(t)=\dfrac{1}{2} kt^{2}\:m/s}}}}}$

Answer 2

GiveN :

• Acceleration is given as a = kt
• Where k is constant
• Initial velocity (u) = 0 m/s

To FinD :

• Velocity in terms of t

SolutioN :

Take the given equation :

$\implies \rm{a \: = \: kt \: \: \: \: \: \: \: \: ...(1)}$

As we know that,

$\implies \rm{a = \dfrac{dv}{dt} \: \: \: \: \: \: \: \: \: ...(2)}$

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Put value of a from (1) in (2)

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$\implies \rm{kt = \dfrac{dv}{dt}}$

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$\implies \rm{dv = kt.dt}$

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Integrate both the sides.

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$\implies \rm{\int dv = \int (kt).dt}$

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$\implies \rm{v = k \int dt^2}$

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$\implies \rm{v = k \dfrac{1}{2} t^2 \: + \: c}$

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$\implies \rm{v \: = \: \bigg(\dfrac{1}{2} \: kt^2 \: + c \: \bigg) \: ms^{-1}}$

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If Integration constant = 0. Then,

$\implies \rm{v \: = \: \dfrac{1}{2} kt^2 \: ms^{-1}}$