205_o6_05 How long does it take a rock falling from rest to reach a velocity

[karush]

$\tiny{205_o6_05_velocity}$
The equation for a free fall at the surface of celestial body in outer space ($s$ in Meters, $t$ in seconds.)\\ is $\displaystyle s=2.38t^2$
How long does it take a rock falling from rest to reach a velocity of $\displaystyle 25.7 \frac{m}{sec}$ on this celestial body in outer space?
$s'=4.76t \\$
$25.7=4.76t\\$
$5.4sec\approx t \\$
how do you do this by $\d{s}{t}$ since s can be confused with seconds?

 

[HallsofIvy]

The problem says "s is in meters" so, no, anyone who was carefully reading the problem would not confuse s with seconds!

 

[karush]

$s=2.38t^2$
$\d{s}{t}=4.76t\d{m}{t}$ ?

 

[Joppy]

$s = displacement$
$v = velocity$
$a = acceleration$

Now, $s' = v \ (in \ m/s)$, and $s'' = a \ (in \ m/s^2)$, also, $v' = a$. We often go back and forth between these three, be it via differentiation or integration. It's always important to check your units though, as you are doing :).

As HallsofIvy has said, it should be obvious to the reader when 's' is used to represent the displacement, and when it is used to represent seconds. To rid yourself of any ambiguity, just do what you have already done and write 'sec'.

 

[karush]

My main question was how is this shown and solved in
Leibniz notation ??

 

[Joppy]

The notation doesn't change the way in which a problem is solved. It is merely a representation of symbols.

The only thing that changes would be your first line i guess.

$\frac{ds}{dt} = v = 4.76t$