
The initial velocity of a particle is \[u\left( {{\text{at }}t = 0} \right)\] and the acceleration is given by \[f = at\]. Which of the following relations is valid?
A. \[v = u + a{t^2}\]
B. \[v = u + \dfrac{{a{t^2}}}{2}\]
C. \[v = u + at\]
D. \[v = u\]
Answer
578.7k+ views
Hint: First of all, write the formula of Newton's second law of motion and equate it to the given equation. Then integrate on both sides by taking their initial and final points to obtain the required answer.
Complete step by step answer:
Formulae used: \[f = \dfrac{{dv}}{{dt}}\]
Given \[f = at................................................\left( 1 \right)\]
From Newton's second law of motion, we have \[f = \dfrac{{dv}}{{dt}}....................................................\left( 2 \right)\]
From equations (1) and (2), we have
\[
\Rightarrow \dfrac{{dv}}{{dt}} = at \\
\Rightarrow dv = atdt \\
\]
Integrating on both sides by taking the limits of initial velocity as \[u\], final velocity as \[v\] and at initial time 0 sec, final time \[t\] sec we get
\[
\Rightarrow \int\limits_u^v {dv} = \int\limits_0^t {atdt} \\
\Rightarrow \left[ v \right]_u^v = a\int\limits_0^t {tdt} \\
\Rightarrow \left[ {v - u} \right] = a\left[ {\dfrac{{{t^2}}}{2}} \right]_0^t{\text{ }}\left[ {\because \int {xdx = \dfrac{{{x^2}}}{2}} } \right] \\
\Rightarrow v - u = a\left[ {\dfrac{{{t^2}}}{2} - \dfrac{0}{2}} \right] \\
\Rightarrow v - u = \dfrac{{a{t^2}}}{2} \\
\therefore v = u + \dfrac{{a{t^2}}}{2} \\
\]
Thus, the correct option is B. \[v = u + \dfrac{{a{t^2}}}{2}\]
Note: Initial velocity is the velocity which the body has in the beginning of the given time period and final velocity is the velocity which the body has at the end of the given time period. Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path a constant speed, it is still accelerating, because the direction of its velocity is changing. Newton's second law of motion describes the effect of net force and mass upon the acceleration of an object.
Complete step by step answer:
Formulae used: \[f = \dfrac{{dv}}{{dt}}\]
Given \[f = at................................................\left( 1 \right)\]
From Newton's second law of motion, we have \[f = \dfrac{{dv}}{{dt}}....................................................\left( 2 \right)\]
From equations (1) and (2), we have
\[
\Rightarrow \dfrac{{dv}}{{dt}} = at \\
\Rightarrow dv = atdt \\
\]
Integrating on both sides by taking the limits of initial velocity as \[u\], final velocity as \[v\] and at initial time 0 sec, final time \[t\] sec we get
\[
\Rightarrow \int\limits_u^v {dv} = \int\limits_0^t {atdt} \\
\Rightarrow \left[ v \right]_u^v = a\int\limits_0^t {tdt} \\
\Rightarrow \left[ {v - u} \right] = a\left[ {\dfrac{{{t^2}}}{2}} \right]_0^t{\text{ }}\left[ {\because \int {xdx = \dfrac{{{x^2}}}{2}} } \right] \\
\Rightarrow v - u = a\left[ {\dfrac{{{t^2}}}{2} - \dfrac{0}{2}} \right] \\
\Rightarrow v - u = \dfrac{{a{t^2}}}{2} \\
\therefore v = u + \dfrac{{a{t^2}}}{2} \\
\]
Thus, the correct option is B. \[v = u + \dfrac{{a{t^2}}}{2}\]
Note: Initial velocity is the velocity which the body has in the beginning of the given time period and final velocity is the velocity which the body has at the end of the given time period. Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path a constant speed, it is still accelerating, because the direction of its velocity is changing. Newton's second law of motion describes the effect of net force and mass upon the acceleration of an object.
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