Kinematics of a point in rectilinear motion

In summary, the problem involves a point Q in rectilinear motion passing through the origin at t=0 and experiencing an acceleration of 6 ft/s^2 to the right until t=5s, then experiencing an acceleration of 12t ft/s^2 to the left. At t=7s, Q is 13 feet to the right of the origin and we need to find the velocity at t=0. The correct equation should be x''=-12t, x'=-6t^2+v1, and x=-2t^3+(v1)(t)+x1. The incorrect equation used the same constants of integration for both the motion from t=5s to t=7s and from t
  • #1
xzibition8612
142
0

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 = 5s, 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


x''=a
x'=at+c1
x=(a/2)t^2+(c1)(t)+c2


The Attempt at a Solution



At t=0:
x''=6
x'=6t+v0
x=3t^2+(v0)(t)

At t=5:
x''=-12t
x'=-6t^2+v1
x=-2t^3+(v1)(t)+x1

At t=7:
x''=-12t
x'=-6t^2+v1
13=-2t^3+(v1)(t)+x1

Now plug in t=5 s into the first set of equations and get:
75+(5)(v0)=x1
30+v0=v1

Plug these two equations into 13=-2t^3+(v1)(t)+x1
As for t in this equation, plug in 2 seconds because this new equation started at t=5 (and hence assume that to be the new starting point, thus 7-5=2).

I get v0 = -15 m/s
The book says the answer is +2. Obviously something went very wrong. Any help would be appreciated.
 
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  • #2
xzibition8612 said:
Now plug in t=5 s into the first set of equations and get:
75+(5)(v0)=x1
30+v0=v1

How did you get the highlighted equation ?
 
  • #3
plug t=5 into x'=6t+v0
Then knowing the fact that at t=5 the velocity is described by x'=-6t^2+v1, I set t=0 at this instant and thus 30+v0 = v1
 
  • #4
you are using same constants on integration in different equations. the motion from t=5 to t=7
should have separate constants of integration from the motion t=7 onwards. that's why you are
mixing equations and got the wrong answer
 
  • #5
But according to the problem, the motion at t=7s follows the same equation as the motion at t=5s (x''=-12), so shouldn't it be the same equations? Can you show me the correct equation? Because from the way I'm reading the problem my way to me is correct, when in fact it's not. I can't see it.
 

Related to Kinematics of a point in rectilinear motion

1. What is kinematics?

Kinematics is the study of motion, specifically the position, velocity, and acceleration of objects.

2. What is rectilinear motion?

Rectilinear motion is motion in a straight line, without any changes in direction.

3. What are the three equations of motion in rectilinear motion?

The three equations of motion in rectilinear motion are:

  1. Position equation: x = x0 + v0t + 1/2at2
  2. Velocity equation: v = v0 + at
  3. Acceleration equation: a = (v - v0)/t

4. How do you find the position, velocity, and acceleration of an object in rectilinear motion?

To find the position, velocity, and acceleration of an object in rectilinear motion, you can use the three equations of motion and plug in the known values for time, initial position, initial velocity, and acceleration.

5. What are some real-life examples of rectilinear motion?

Some real-life examples of rectilinear motion include a car moving in a straight line on a highway, a ball falling from a height, and a person running in a straight line.

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