Zenoture
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Physics: physical representation of the math
r(t) is the position function of a particle moving in 2 or 3 space.
**I don't know how to do integral symbol, so I'll just denote that symbol as integ(f(x))**
integ( ||dr/dt|| ) from t0 to t1
I know dr/dt = v(t) which is the velocity function, and the norm of v(t) is the magnitude of the vector, aka, the numerical speed of the particle at time t, but when we integrate the speed, wouldn't we just end up with the position function again?
That is what my intuition is telling me, but my intuition has failed me before, so I would rather be shown why my thinking is wrong, if it even is. Thanks.
**I don't know how to do integral symbol, so I'll just denote that symbol as integ(f(x))**
integ( ||dr/dt|| ) from t0 to t1
I know dr/dt = v(t) which is the velocity function, and the norm of v(t) is the magnitude of the vector, aka, the numerical speed of the particle at time t, but when we integrate the speed, wouldn't we just end up with the position function again?
That is what my intuition is telling me, but my intuition has failed me before, so I would rather be shown why my thinking is wrong, if it even is. Thanks.
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If I were to assume some random value for || v(t) || to be 10 m/s, upon integration I would end up with 10t, where t is in seconds, and 10 is still in meters per second. So by simple plug and chug, where t = 1...
10 m/s * 1 s = 10m therefore proving that the above statement is the displacement of the particle after t seconds.
Right?