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liechti1
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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.
(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?
[tex]
v = \frac{ds}{dt}
[/tex][tex]
a = \frac{dv}{dt}
[/tex]For constant acceleration:[tex]
v = v_o + at
[/tex][tex]
s = s_o + v_ot + \frac{1}{2}at^2
[/tex][tex]
v^2 = v_o^2 + 2a(s - s_o)
[/tex]
A few notes:
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!
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
[tex]
v = \frac{ds}{dt}
[/tex][tex]
a = \frac{dv}{dt}
[/tex]For constant acceleration:[tex]
v = v_o + at
[/tex][tex]
s = s_o + v_ot + \frac{1}{2}at^2
[/tex][tex]
v^2 = v_o^2 + 2a(s - s_o)
[/tex]
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).
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!
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