Question

algorithm to obtain the kinematics equation for uniform acceleration.​

Answer 1

Explanation:

$s = ut + \dfrac{1}{2}a{t^2}$ C. ${v^2} = {u^2} + 2as$ Hint: All the three equations for the uniformly accelerated motion can be derived by using the expressions for velocity, displacement, and acceleration and integrating them.

Answer 2

To obtain : algorithm for kinematics equation for uniform acceleration.​

Solution :

Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:

• v = u + at

       Acceleration = Change in velocity/Time Taken

       Therefore,  Acceleration = (Final Velocity-Initial Velocity) / Time Taken

       Hence, a = v-u /t or at = v-u

       Therefore, we have:

        v = u + at

• v² = u² + 2as

      We have, v = u + at. Hence, we can write t = (v-u)/a  

      Also, we know that, Distance = average velocity × Time

       Therefore, for constant acceleration we can write:

      Average velocity = (final velocity + initial velocity)/2 = $\[\frac{{v + u}}{2}\]$

        Hence, Distance (s) = [($\[\frac{{v + u}}{2}\]$]  × [($\[\frac{{v - u}}{a}\]$)]

        or  s = $\[\frac{{{v^2} - {u^2}}}{{2a}}\]$

       or 2as = v² – u²

        v² = u² + 2as

•   s = $\[ut + \frac{1}{2}a{t^2}\]$

      Let the distance be “s”.

       Distance = Average velocity ×  Time.

      Also, Average velocity = $\[\frac{{u + v}}{2}\]$

      Therefore, Distance (s) = $\[\frac{{u + v}}{2}\]$ × t

       Also, from v = u + at, we have:

       s = $\[\frac{{u + u + at}}{2}\]$ × t =$\[\frac{{2u + at}}{2}\]$× t

       s =$\[\frac{{2ut + a{t^2}}}{2}\]$ = $\[\frac{{2ut}}{2} + \frac{{a{t^2}}}{2}\]$

        s = $\[ut + \frac{1}{2}a{t^2}\]$

Hence we can obtain the kinematics equation for uniform acceleration from following algorithm and expression.​