Definite Integrals: Solving an Easy Integral with Acceleration a=2 m/s^2

[macarroni]

I don't know how to solve this really easy integral.

suppose: a=2 m/s^2

a = dv / dt

\int^b_a a \ dt = \int^b_a dv

\int^b_a 2 \ dt = v_{b} - v_{a}

2t_{b} - 2t_{a} = v_{b} - v_{a}

v_{b} = v_{a} + 2t_{b} - 2t_{a}

I hope everything is ok up to this point. Then I want to know the position: r, and I need to use the same b and a that I have already use before, but...

v = dr / dt

\int^b_a v \ dt = \int^b_a dr

\int^b_a ( v_{a} + 2t_{b} - 2t_{a} ) \ dt = r_{b} - r_{a}

v_{a}t_{b} - v_{a}t_{b} ... \ ? = r_{b} - r_{a}

 

[James R]

In your first expression for a, is m the mass and s the displacement? Or is s just a constant?

 

[StatusX]

It looks like you're trying to find the velocity and position of an object under accelaeration 2 m/sec^2 at a time t_b in terms of its velocity at some earlier time t_a. If so, you found v_b correctly. But to get the position, you need to integrate the velocity as a function of time, not the velocity at t_b. To get v(t), just replace t_b with the variable t.

 

[Gamma]

a = 2 m/s^2

First part is OK. But when you did the second part you need a general form of v to do the diferentiation. This is how I would do this:

d^2 r/dt^2 = a

Integrating once, you have

dr/dt = a t + A --------------------(1)

where A is a constant and we can find A using the boundary conditions. You have been given at time t = ta , v = va.

so, v_a = a t_a + A => A = v_a - a t_a

so v(t) = at + ( v_a - a t_a)

Integrate (1) wrt t

r = at^2/2 + At + B -----------------(2)

where B is a constant and can be found as earlier using Boundary conditions. at t=ta, r = ra .

r_a = at_a^2/2 + At_a + B

Solve for B and subtitute in (2) to get an expression for r(t)

 

[xanthym]

.


Because velocity "v" is a function of time "t", which is indicated by writing it "v(t)", you need to maintain the functional dependence of v(t) into the second set of integrations. Your original work is rewritten below in a manner to do that. Also, equations become more managable if {t_a=0}.

QUESTION: What is the distance "r(t)" traveled from time t=0 to time "t" (for constant acceleration "a")??

a \ = \ dv/dt

\int^{t}_{0} a \ dt \ = \ \int^{v(t)}_{v(0)} \ dv

at \ = \ v(t) - v(0)

\color{blue} v(t) \ = \ v(0) + at


v \ = \ dr/dt

\int^{t}_{0} v(t) \ dt \ = \ \int^{r(t)}_{r(0)} \ dr

\int^{t}_{0} \color{blue}\left( v(0) + at \right)\color{black} \ dt \ = \ r(t) - r(0)

v(0)t + (1/2)at^{2} \ = \ r(t) - r(0)

\color{red} r(t) \ = \ r(0) + v(0)t + (1/2)at^{2}

Now place {a=(2 m/sec^2)} and values for "r(0)" and "v(0)" into the equation for the final result.


~~

 

[Nylex]

In your first expression for a, is m the mass and s the displacement? Or is s just a constant?

. Those are units..

 

[James R]

Those are units..

Ah! I completely misread. It's a = 2 metres per second squared. Silly me! I think I just assumed that since the rest of the post was in terms of variables, the acceleration was in terms of variables, too.

I'd do this similarly to how Gamma did it, above.