Rectilinear Motion: Solving for t=0 Velocity

• xzibition8612
In summary, at t=0, point Q was at the origin and had a velocity of 6 ft/s. At t=5 seconds, the velocity of Q had increased to 12 ft/s to the right. At t=7 seconds, the velocity of Q had increased to 13 ft/s to the right.
xzibition8612

Homework Statement

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?

The Attempt at a Solution

Assume right is positive and left is negative.

t=0 to t=5
x''=6
x'=6t+(V0)
x=3t^2+(V0)t

From t=5 onwards
x''=-12t
x'=-6t^2+(V1)
x=-2t^3+(V1)t+[75+5(V0)t] (parts in brackets [] obtained by using the x=3t^2+(V0)t equation setting t=5)

we are given 13=-2t^3+(V1)t+[75+5(V0)t] at t=7s (from beginning) or t=2s (considering the equation changes at t=5)

thus plugging in t=2 (since x''=-12t begins at t=5) get:

-46=2(V1)+10(V0)

Now need another equation to solve variables. Equating the equations x'=6t+(V0) (t=0 to t=5) and the equation x'=-6t^2+(V1) (t=5 onwards)

Thus plug in t=5 for x'=6t+(V0) and t=0 for x'=-6t^2+(V1) and then equate them to get

30+(V0)=V1

Solve and get V0 = -8.83 ft/s

Book says answer is 2ft/s to the right.

Something went terribly wrong...

hi xzibition8612!

hmm … this is exactly why you should never take short-cuts
xzibition8612 said:
t=0 to t=5
x''=6
x'=6t+(V0)
x=3t^2+(V0)t

From t=5 onwards
x''=-12t
x'=-6t^2+(V1)
x=-2t^3+(V1)t+[75+5(V0)t] (parts in brackets [] obtained by using the x=3t^2+(V0)t equation setting t=5)

that 5(V0)t should be 5V0, shouldn't it?

in between those two paragraphs, you should have paused to state explicitly the final values of x and v, that you intend to use later​

Yuo're right. But then I get 2(30+V0)+5(V0)=-46

Hence V0 = -15 m/s

which is still wrong. No idea where i went wrong.

hi xzibition8612!

it's very difficult to check your work

(i don't see a 73 anywhere)

can you write it out in one continuous proof?​

tiny-tim did you get 2 ft/s as the correct answer? If you did, can you just show me how to do it? Because I'm afraid its very difficult to communicate online and if I see what you did then I can figure out where I went wrong thanks.

1. What is rectilinear motion?

Rectilinear motion is the motion of an object in a straight line, with constant velocity and no acceleration.

2. How do you solve for t=0 velocity in rectilinear motion?

To solve for t=0 velocity in rectilinear motion, you can use the formula v = d/t, where v is the velocity, d is the distance, and t is the time. Rearrange the formula to solve for v, and plug in the values for d and t.

3. What is the importance of finding t=0 velocity in rectilinear motion?

Finding t=0 velocity in rectilinear motion is important because it helps us understand the starting velocity of an object, which is necessary for predicting its future motion and determining its acceleration.

4. Can you use the same method to solve for t=0 velocity in objects with varying acceleration?

Yes, the formula for t=0 velocity can also be used for objects with varying acceleration. However, in this case, the velocity will only be accurate for the specific moment in time when t=0.

5. Are there any limitations to using t=0 velocity in rectilinear motion?

Yes, there are limitations to using t=0 velocity in rectilinear motion. This method assumes that the object is already in motion at t=0 and does not account for any external forces that may affect the motion of the object. It is best used for objects with constant velocity and no acceleration.

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