
The initial velocity of a particle is u (at t=0), and acceleration is given by $f=at$. Which of the following relations are valid?
A. $v=u+a{{t}^{2}}$
B. $v=u+\dfrac{a{{t}^{2}}}{2}$
C. $v=u+at$
D. $v=u$
Answer
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Hint: We have a particle whose velocity is given by u at t=0, and its acceleration is a linear function of time. We know that the rate of change of velocity is the acceleration of a body. So, we should apply this relation here for the given acceleration and integrate it to get the velocity of the particle at time t.
Complete step by step answer:
In the problem, the particle has an initial velocity u at time t=0. Its acceleration is a linear function of time given by $f=at$, this means acceleration is not a constant and it changes with time. We know that the acceleration of a body is defined as the rate of change of velocity. So, we can write,
$a=\dfrac{dv}{dt}$
In the case of our problem, we can write the above equation as,
$\dfrac{dv}{dt}=f=at$
$\Rightarrow dv=\left( at \right)dt$
Integrating the above equation with u as the initial velocity at time t=0 and v as the final velocity at a time t, we can write,
$\int\limits_{u}^{v}{dv}=\int\limits_{0}^{t}{\left( at \right)dt}$
Integrating the above equation and applying the limits, we get,
$v-u=\dfrac{a{{t}^{2}}}{2}$
$\therefore v=u+\dfrac{a{{t}^{2}}}{2}$
So, the answer to the question is an option (B).
Note: Newton’s law of motion is defined for motion is only valid for the motion of a body having constant acceleration, that is the acceleration of the body should not change with time. In the given problem, the body does not have a constant acceleration, and it varies linearly with time.
The velocity of a body is defined as the rate of change of displacement of the body.
The slope of the displacement time graph gives the velocity of the body, while the slope of the velocity-time graph gives the acceleration of the body.
The area under the velocity-time graph gives the displacement of the body in that period of time. While the area under the acceleration-time graph gives the velocity of the body in that period of time.
Complete step by step answer:
In the problem, the particle has an initial velocity u at time t=0. Its acceleration is a linear function of time given by $f=at$, this means acceleration is not a constant and it changes with time. We know that the acceleration of a body is defined as the rate of change of velocity. So, we can write,
$a=\dfrac{dv}{dt}$
In the case of our problem, we can write the above equation as,
$\dfrac{dv}{dt}=f=at$
$\Rightarrow dv=\left( at \right)dt$
Integrating the above equation with u as the initial velocity at time t=0 and v as the final velocity at a time t, we can write,
$\int\limits_{u}^{v}{dv}=\int\limits_{0}^{t}{\left( at \right)dt}$
Integrating the above equation and applying the limits, we get,
$v-u=\dfrac{a{{t}^{2}}}{2}$
$\therefore v=u+\dfrac{a{{t}^{2}}}{2}$
So, the answer to the question is an option (B).
Note: Newton’s law of motion is defined for motion is only valid for the motion of a body having constant acceleration, that is the acceleration of the body should not change with time. In the given problem, the body does not have a constant acceleration, and it varies linearly with time.
The velocity of a body is defined as the rate of change of displacement of the body.
The slope of the displacement time graph gives the velocity of the body, while the slope of the velocity-time graph gives the acceleration of the body.
The area under the velocity-time graph gives the displacement of the body in that period of time. While the area under the acceleration-time graph gives the velocity of the body in that period of time.
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