Answer:
- 32.17 fts/s in the imperial system, -9.8 m/s in the SI
Explanation:
We know that acceleration its the derivative of velocity with respect to time, this is (in 1D):
[tex]a = \frac{dv}{dt}[/tex]
So, if we wanna know the change in velocity, we can take the integral:
[tex]v(t_f) - v(t_i) = \int\limits^{t_f}_{t_i} {\frac{dv}{dt} \, dt = \int\limits^{t_f}_{t_i} a \, dt[/tex]
Luckily for us, the acceleration in this problem is constant
[tex]a \ = \ - g[/tex]
the minus sign its necessary, as downward direction is negative. Now, for a interval of 1 second, we got:
[tex]t_f = t_i + 1 s[/tex]
[tex]v(t_i + 1 s) - v(t_i) = \int\limits^{t_i + 1 s}_{t_i} a \, dt[/tex]
[tex]v(t_i + 1 s) - v(t_i) = \int\limits^{t_i + 1 s}_{t_i} (-g ) \, dt[/tex]
[tex]v(t_i + 1 s) - v(t_i) = [-g t]^{t_i + 1 s}_{t_i} [/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g (t_i+1s) + g t_i [/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g t_i + -g 1s + g t_i[/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g 1s [/tex]
taking g in the SI
[tex]g=9.8 \frac{m}{s^2}[/tex]
this is:
[tex]v(t_i + 1 s) - v(t_i) = - 9.8 \frac{m}{s} [/tex]
or, in imperial units:
[tex]g=32.17 \frac{fts}{s^2}[/tex]
this is:
[tex]v(t_i + 1 s) - v(t_i) = - 32.17 \frac{fts}{s} [/tex]
