Question

The initial velocity of a particle is $u$ (at t=0) and the acceleration $a$ is given by$ft$. Which of the following relations is valid?
A.) $v = u + f{t^2}$
B.) $v = u + \dfrac{{f{t^2}}}{2}$
C.) $v = u + ft$
D.) $v = u$

Answer 1

Hint: For answering these types of questions we must remember the equation of motions. There are 3 equations of motion.

Complete step-by-step answer:

Equations of motion are equations which describe the behavior of a physical system as a function of time in terms of its motion.

Here $a$ is the acceleration which is given by $ft$ and $t$ is the time.

Therefore, acceleration $a$=$\dfrac{{dv}}{{dt}}$,
$a$=$ft$

Acceleration is the rate of change of velocity,

Hence it can also be written as $a$=$\dfrac{{dv}}{{dt}}$, where $v$ is the velocity and $t$is the time.

$a$=$\dfrac{{dv}}{{dt}}$
$ft$=$\dfrac{{dv}}{{dt}}$
$dv = ft \cdot dt$

Integrating both the sides, we get

$\int\limits_u^v {dv} = \int\limits_0^t {ft \cdot dt}$
$v - u = \dfrac{{f{t^2}}}{2}$
$v = u + \dfrac{{f{t^2}}}{2}$

Therefore, our option B is the correct answer. Which states that is initial velocity of a particle is $u$ (at t=0) and the acceleration $a$ is given by$ft$then $v = u + \dfrac{{f{t^2}}}{2}$ is the correct relation.

Note: Three equations of motions are $v = u + at$, $S = ut + \dfrac{1}{2}a{t^2}$ and ${v^2} = {u^2} + 2aS$ where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $t$ is the time and $S$ is the distance covered. These equations are very important and we must remember this always.