Question

Example 83 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

Answer:

Explanation:

a=kt where k is constant to be determined. Now, a=dv/dt=kt. Integrating over t,we get

v=(1/2)kt^2+c1, where c1 is constant of integration. Using initial condition that at t=0,v=0 we have c1=0.

Then, v=(1/2)kt^2

Answer 2

Given, the acceleration,

$\longrightarrow\sf{a(t)=kt}$

But we know that acceleration is the first derivative of velocity with respect to time.

$\longrightarrow\sf{a(t)=\dfrac {d}{dt}[v(t)]}$

Then,

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

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

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

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

$\displaystyle\longrightarrow\sf{v(t)=k\cdot\dfrac {t^2}{2}+c}$

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

Since the motion starts from rest, $\displaystyle\sf{v=0}$ at $\displaystyle\sf{t=0.}$ Thus,

$\displaystyle\longrightarrow\sf{v(0)=0}$

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

$\displaystyle\longrightarrow\sf{c=0}$

Therefore,

$\displaystyle\longrightarrow\sf{\underline {\underline {v(t)=\dfrac {1}{2}kt^2}}}$