Question

For motion in a straight line with constant acceleration, we derived two primary equations given by Eq. (v = u + at) and (s = ut + `1/2 xx at^2`​). Using these two equations, three more equations can be derived, out of which we derived one given in Eq. (v² = u² + 2as). Derive the remaining two equations given below

`s = vt - 1/2 at^2              s = 1/2 (u + v)t`

In mathematics, you have learnt the formula for calculating the area of a trapezium. Using that formula, derive the second equation given above.

Answer 1

The two equations can be derived using the first two equations of motion:

v = u + at

s = ut + `1/2 xx at^2`​

1. Derive `s = vt - 1/2 at^2`

From the first equation:

u = v − at

Substitute this into

`s = ut + 1/2 at^2`

`s = (v - at)t + 1/2at^2`

= `vt - at^2 + 1/2 at^2`

= `vt - 1/2at^2`

Hence,

`s = vt - 1/2at^2`

2. Derive `s = 1/2 (u + v)t`

Using the trapezium area method:

Draw a velocity-time graph.

Initial velocity = u

Final velocity = v

Time = t

The graph forms a trapezium.

The area under the velocity-time graph gives the displacement.

Area of a trapezium:

Area = `1/2 xx ("sum of parallel sides") xx ("height")`

Here,

Parallel sides = u and v

Height = t

Therefore,

`s = 1/2 (u + v) xx t`

= `1/2 (u + v)t`

Hence,

`s = 1/2 (u + v)t`