Question

Show that for motion in a straight line with constant acceleration \(a\), initial velocity \({v_0}\), and initial displacement \({s_0}\), the displacement after time \(t\) is

\(s = \frac{1}{2}a{t^2} + {v_0}t + {s_0}\)

Short answer

It is proved that \(s = \frac{1}{2}a{t^2} + {v_0}t + {s_0}\).

Step-by-step solution

Finding velocity

Here we have given that acceleration is constant.

So,\(a\left( t \right) = a,\;\;\;\;a{\rm{ is any constant}}\)

Now, we know that velocity is anti-derivative of acceleration.

So, the general anti-derivative of\(a\left( t \right) = a,\;\;\;\;a{\rm{ is any constant}}\)is,

\(v\left( t \right) = at + {v_0},\;\;\;\;a,\;{v_0}\left( {{\rm{initial velocity}}} \right){\rm{ are any constant}}\).

Prove that \(s = \frac{1}{2}a{t^2} + {v_0}t + {s_0}\)

Here we have,\(v\left( t \right) = at + {v_0},\;\;\;\;a,\;{v_0}\left( {{\rm{initial velocity}}} \right){\rm{ are any constant}}\)

We know that position is anti-derivative of velocity.

So, take anti-derivative of\(v\left( t \right) = at + {v_0}\).

\(s\left( t \right) = a\frac{{{t^2}}}{2} + {v_0}t + {s_0},\;\;\;\;a,\;{v_0}\left( {{\rm{initial velocity}}} \right){\rm{,}}{{\rm{s}}_0}\left( {{\rm{initial position}}} \right){\rm{ are any constant}}\)

So,\(s = \frac{1}{2}a{t^2} + {v_0}t + {s_0}\).

Hence proved.