Question

For a freely falling body is velocity taken negative

Answer 1

S = ut + 12 at2

Where,

S = displacement

u = initial velocity

a = acceleration (must be constant),

t = time.

This equation DOES NOT relate to final velocity.

Analytical Proof

We know that,

v = u + at

Also,

v = dSdt

‘S’ is the displacement.

Therefore, equating both,

dSdt = u + at

S∫0 dS = t∫0 (u + at) . dt

S = ut∫0 dt + a t∫0 t . dt

S = ut + 12 at2

Graphical Proof

Following is a v - t graph displaying constant acceleration. (Slope of the curve is constant)



At t = 0 seconds, the particle’s velocity is u m/s.

At t = t seconds, the particle’s velocity is v m/s.

Area under the curve of v − t graph gives displacement.

Now,

S = Area of Rectangle + Area of Triangle

S = (u - 0) × (t - 0) + 12(v - u)(t - 0)

S = ut + 12(v - u)(t)

(Substituting v − u = at, from first equation of motion and rearranging the terms)

S = ut + 12 at2