Understanding velocity and acceleration

[coverband]

speed = distance/time
v=s/t
Acceleration = dv/dt = -s/t^2 ?

 

[andrewkirk]

In the equation v = s/t, 'v' represents average speed over the period from time 0 to time t.

In the equation accel = dv/dt, 'v' represents instantaneous speed at time t.

Unless acceleration is always zero, average speed and instantaneous speed are different items, and cannot be represented by the same variable letter ('v' in this case).

 

[DrGreg]

Another thing to add is that the differentiation in post #1 would be valid only if $s$ were a constant and $t$ were a variable. Does that make sense?

 

[coverband]

Another thing to add is that the differentiation in post #1 would be valid only if $s$ were a constant and $t$ were a variable. Does that make sense?

No because when deriving an equation for v we start with a = dv/dt -> dv = a dt -> v=[int]a dt -> v = u + at. This is how Wikipedia derives the first equation of motion. They treat a as a constant. Thanks for your first answer though

 

[DrGreg]

No because when deriving an equation for v we start with a = dv/dt -> dv = a dt -> v=[int]a dt -> v = u + at. This is how Wikipedia derives the first equation of motion. They treat a as a constant. Thanks for your first answer though

I said constant $s$, not constant $a$. My comments refer specifically to $$
\frac{d}{dt} \left( \frac{s}{t} \right) = -\frac{s}{t^2},
$$which is valid only if $s$ is constant.