- #1

wahaj

- 156

- 2

If an object is dropped from rest the acceleration is given by

[tex] a = 9.81(1 - \frac {v^2}{10000}) [/tex]

what is velocity after 7 seconds.

I know that [tex] a = \frac{dv}{dt}[/tex]

so [tex] dv = a \cdot dt [/tex]

here I chose time as the variable I am integrating with respect to because I am given time in my question.

[tex] \int^v_0 dv = \int^t_09.81(1 - \frac {v^2}{10000}) [/tex]

[tex] v = 9.81t (1 - \frac {v^2}{10000}) [/tex]

solving the quadratic at t = 7 I get v = 50.9

but the book integrates with respect to v as in

[tex] dt = \frac{dv}{a} [/tex]

putting in the values and solving this I get 59.6 as my answer which is correct.

What confuses me is the choice between whether to isolate dv and integrate w.r.t dt or the other way around. Same problem arises when I am using other equations especially

[tex] a = v \frac{dv}{ds}[/tex]

Is there some sort of rule or trick to figuring this out because my math class is way behind my dynamics class and my dynamics class isn't going to slow down.