The problem involves calculating the position vector of a particle given its velocity vector, which includes a term indicating acceleration. Participants express confusion regarding the units and the implications of the acceleration term in the context of velocity.
The discussion includes attempts to clarify the relationship between the components of the vectors and the constants of integration. Some participants suggest considering both definite and indefinite integrals, while others note the need for initial conditions to determine specific values.
Participants mention the context of a take-home test, which may influence their approach to providing answers. There is also a reference to the need for specific values to evaluate definite integrals.
brinstar said:Homework Statement
Calculate the position vector of a particle moving with velocity given by:
v = (32 m/s - (5 m/s^2 )t i) + (0 j)
Homework Equations
(x^(n+1) / (n+1) ) + C = antiderivative of function
The Attempt at a Solution
r = (32t m - (5/2)t^2 m/s + C m i) + (C j)
Honestly, I'm just confused with the units more than anything. I don't know why the problem has m/s^2 if it's a velocity vector...
SteamKing said:A quantity of 5 m/s2 indicates that accelerated motion is taking place, i.e., the velocity is changing w.r.t. time. Acceleration × time = change in velocity.
I would say that since the velocity vector had no j-component, the position vector will not either.brinstar said:oooooh okay that makes more sense. thank you!
and the antiderivative is right, right?
but isn't the antiderivative of 0 C (or in this case, D to differentiate)?SteamKing said:I would say that since the velocity vector had no j-component, the position vector will not either.
Yeah, but D = 0 would be an acceptable value for the constant of integration, in the absence of any other initial condition information.brinstar said:but isn't the antiderivative of 0 C (or in this case, D to differentiate)?
SteamKing said:Yeah, but D = 0 would be an acceptable value for the constant of integration, in the absence of any other initial condition information.
You can have a definite integral only if you know the value of t.brinstar said:oh okay. since this is on a take home test, do you think I should just put both answers (one for a definite integral 0 and one for an indefinite integral D)?
SteamKing said:You can have a definite integral only if you know the value of t.