Physics Velocity Equation, t, t-prime in v(t)=v_0 + int[t_0,t] a(t') dt'

[prepforcalc]

Why put primes on some of these t's but not on all of them? Can you give an example of where t is unequal to t prime in this case?

. . . . .\(\displaystyle \displaystyle v(t)\, =\, v_0\, +\, \int_{t_0}^t\, a(t')\, dt'\)

 

[tkhunny]

t' is a dummy variable to make the integral serviceable. Change each t' to "z" and see if you feel better about it.

I suspect the author wishes to emphasize that we are still talking about 't', both inside and outside of the integral. Notation dictates that they are not known by the same name.

 

[Deleted member 4993]

Why put primes on some of these t's but not on all of them? Can you give an example of where t is unequal to t prime in this case?

. . . . .\(\displaystyle \displaystyle v(t)\, =\, v_0\, +\, \int_{t_0}^t\, a(t')\, dt'\)

Here t' is known as dummy variable - i.e. - after integration and taking appropriate limits (for definite integral) - no trace of t' will remain in your answer.

Some books prefer to use Θ in this case.

 

[prepforcalc]

The context is a graph with acceleration on the y axis, and time on the x axis. The x axis is labeled, 0, to, t1, t2, t3, t4, t', t. Then there is a series of questions about whether the velocity or acceleration of a particle is increasing, decreasing etc. at a specific point on the graph. Given this information would you still conclude that it's a dummy variable? And if so, in what sense? I didn't post an actual screen shot of the entire question due to copyright, but I can be more specific if that would be helpful.

 

[tkhunny]

Absolutely, it's a dummy variable. You provided no discouraging context. In what sense? In the sense that it serves ONLY to make the integral work and does nothing else - except as t' it is reminiscent of t - the thing it represents inside the integral.