Why is 's' positive?

Question

Given that:

s=Displacement

u=Initial Velocity

v=Final Velocity

a=Acceleration

t=Time

g=Gravitational Field Strength (modeled as 10m/s^2)

Assume lack of air resistance

A particle, 35 metres off of the ground, is projected upwards, and takes 4 seconds to fall to the ground.

The acceleration should therefore be -10, the total net displacement during that time should be -35, and the time is obviously 4 seconds.

Since s=vt - 1/2 at^2

-35=4v - 1/2*-10*4^2

-35=4v+80

-115=4v

v=-28.75

However, according to my teacher, s is positive. I cannot comprehend this, since a is stated to be negative, and it is accelerating towards the ground, and s is a vector in the same direction as the acceleration.

His answer was something along the lines of s being the 'starting value'. Can anyone explain this to me?

EDIT: I put 10 instead of -10 for a

Answer 1

Your teacher used $a=g=10 m/s^2$, so your teacher used down is positive, and with the displacement being down as well, the displacement is therefore positive as well.

In your calculaton, you used $a=-10$, so you used down being negative, and therefore got a negative displacement ($-35$) as well.

In the end, you're both right; you just differed in what direction you considered positive.

Answer 2

It is mainly matter of convention, indeed we can assume any origin in teh space and any direction as positive arranging the equations accordingly.

In this case the teacher has assumed the origin at the ground and positive the upward direction for $s$ and thus the acceleration $a=-g$ has been assumed negative since it is directed downward.

Notably the equation for the motion is

• $s(t)=s_0+v_0t+\frac12at^2=35+v_0t-\frac12gt^2$

with the condition that $s(4)=0$ we have

• $s(4)=35+4v_0-80=0 \implies$ initial velocity $v_0=11.25 m/s$

which is positive accordingly to the convention adopted.