Wait, i think it just hit me as I posted the question.

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?

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?