Kinematics: Rectilinear Motion (calculus)

[liechti1]

Hello, first post here. I've used these forums as reference for quite some time, just never actually made an account until now.

I'm working through the textbook for my upcoming Intro to Dynamics class and I am having trouble figuring out this problem. It seems like it should be pretty straightforward, but perhaps not.

Homework Statement

(from An Introduction to Dynamics, 4th ed. by Mcgill/King)
1.41) A point Q in rectilinear motion passes through the origin at t = 0, and from then until 5 seconds have passed, the acceleration of Q is 6 ft/s^2 to the right. Beginning at t = 5 seconds, the acceleration of Q is 12t ft/s^2 to the left. If after 2 more seconds point Q is 13 feet to the right of the origin, what was the velocity of Q at t = 0?

Homework Equations

$$v = \frac{ds}{dt}$$$$a = \frac{dv}{dt}$$For constant acceleration:$$v = v_o + at$$$$s = s_o + v_ot + \frac{1}{2}at^2$$$$v^2 = v_o^2 + 2a(s - s_o)$$

The Attempt at a Solution

A few notes:

• I'll choose x to be positive in the right direction.
• There are two time intervals. 0 to 5s (subscript 1), and t > 5s (subscript 2).

Here's what the problem gives us:
a₁ = 6 ft/s²
a₂ = -12t ft/s²
s(7) = 13 ft

So I think the main idea is to use the initial condition and a₂ to figure out either the position or velocity at t = 5s. From there, any of the constant acceleration formulas will suffice to find the initial velocity.

a₂ = -12t
v₂ = -6t² + c₁
s₂ = -2t³ + c₁t + c₂

Plugging the initial condition still leaves me with 2 constants. I know that at t = 5s, a₁ = a₂, v₁ = v₂, s₁ = s₂, but I'm not sure how to relate the first time interval to the second.

4. The Answer
The back of the book says: 2 ft/s (to the right)

Any help/tips would be appreciated, thanks!

 

[tedbradly]

Let a_n be the acceleration during the nth part and t_n be the duration that a_n lasted.

$$s_2(t_2) = v_{20} t_2 + x_{20} + \frac{1}{2} a_2 t_2^2$$
$$v_{20} = v_1(t_1) = v_{10} + a_1 t_1$$
$$x_{20} = x_1(t_1) = v_{10} t_1 + \frac{1}{2} a_1 t_1^2$$

Plugging in the last two into the first:

$$s_2(t_2) = (v_{10} + a_1 t_1) t_2 + v_{10} t_1 + \frac{1}{2} a_1 t_1^2 + \frac{1}{2} a_2 t_2^2$$
$$v_{10}=\frac{s_2(t_2) - \frac{1}{2}(a_1 t_1^2 + a_2 t_2^2+2 a_1 t_1 t_2)}{t_2+t_1}$$

 

[gneill]

You'll need to incorporate the given final position of +13 ft at t=7 seconds into the mix. In order to do that you have to find an expression for the distance for t > 5 seconds. That's going to involve a bit of calculus, particularly since it the acceleration depends upon time for t > 5.

The first leg of the journey is simple enough since the acceleration is constant. You can use the usual constant acceleration formulae to determine the displacement and velocity at time t = 5, keeping the initial velocity as a variable of course.

For the second leg, two integrations will be required. The first to find an expression for the change in velocity for this period, and the second to combine this with the velocity left over from the first leg to find the change in distance from the t = 5 epoch.

 

[tedbradly]

I didn't see the t for the second acceleration. In that case, everything is the same, save the equation for s_2(t) (which will have a few t_2s and 1/2 or 1/3 constants floating around in it now from integration). But I'm sure you can use the model I've put in place to dispense of your prior confusion.

 

[liechti1]

Hey, thanks for the advice guys. I am still not quite getting it, here is my work:
$$s_2(t_2) = v_{20} t_2 + x_{20} + \frac{1}{6} a_2 t_2^3$$$$v_{20} = v_1(t_1) = v_{10} + a_1 t_1$$$$x_{20} = x_1(t_1) = v_{10} t_1 + \frac{1}{2} a_1 t_1^2$$
Plugging in the last two into the first:
$$s_2(t_2) = (v_{10} + a_1 t_1) t_2 + v_{10} t_1 + \frac{1}{2} a_1 t_1^2 + \frac{1}{6} a_2 t_2^3$$$$13 = (v_{10} + (6)(5))(2) + 5v_{10} + \frac{1}{2}(6)(5^2) + \frac{1}{6}(-12)(2^3)$$$$13 = 2v_{10} + 60 + 5v_{10} + 75 + (-16)$$$$-106 = 7v_{10}$$$$v_{10} = -15.14$$

The correct answer, according to the back of the book is 2 ft/s. This may be purely coincidental, but I noticed this:

If I use this equation, it works out (notice the - a₁t₁ instead of + a₁t₁):
$$s_2(t_2) = (v_{10} - a_1 t_1) t_2 + v_{10} t_1 + \frac{1}{2} a_1 t_1^2 + \frac{1}{6} a_2 t_2^3$$$$13 = 2v_{10} - 60 + 5v_{10} + 75 + (-16)$$$$14 = 7v_{10}$$$$v_{10} = 2$$
When plugging the numbers into this equation you get v₁₀ = 2 ft/s.

Does the book have a typo? Or is second equation correct? If it is, I do not understand where the negative sign comes from. Or, am I making another error somewhere else?

 

[tedbradly]

It doesn't make sense to me, but I'm not the best. Personally, I believe myself over a textbook any day.

One time on a test with simple convolution, I did it by hand as you're supposed to do. Then, I plugged in the integral into my calculator [the question was asking for the convolution evaluated at a certain argument] and got another answer. I boldly kept mine, receiving full credit, disobeying the electric gods.