One dimensional Kinematics help

In summary, the elevator is ascending with an upward acceleration of 4.0 ft/s2 and at the instant its upward speed is 8.0 ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor. Using the equations y=yinitial+vt+1/2at2 and y=8t+2t2, the time of flight of the bolt from ceiling to floor is calculated to be 1.14 seconds. The distance it has fallen relative to the elevator shaft is found to be 2.75 meters. However, it should be noted that the value of g used in the calculations was incorrect and the definition of y=0 was not accurate, leading to potential
  • #1
toesockshoe
265
2

Homework Statement


2. An elevator ascends with an upward acceleration of 4.0 ft/s2. At the instant its upward speed is 8.0 ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor. Calculate:
a. the time of flight of the bolt from ceiling to floor.
b. the distance it has fallen relative to the elevator shaft.

Homework Equations


y=yinitial+vt+1/2at2

The Attempt at a Solution


assuming that elevator starts at bottom of shaft and y=0 is at bottom of shaft.

a) y value of bolt has to equal the y value of the elevator.
BOLT EQUATION:
y = 9+8t-(1/2)gt^2
FLOOR OF ELEVATOR EQUATION:
y=8t+2t^2

set bolt and floor equation equal to each other to get t = 1.14 seconds.

b) delta y of bolt respect to shaft = 8t -1/2gt^2
plug in 1.14 for t to get 2.75 meter increase?

please tell me if i am going wrong and where...
 
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  • #2
toesockshoe said:
BOLT EQUATION:
y = 9+8t-(1/2)gt^2
FLOOR OF ELEVATOR EQUATION:
y=8t+2t^2
Looks right so far..
toesockshoe said:
t = 1.14 seconds
Too long. Did you try substituting this back in both equations as a check?
 
  • #3
haruspex said:
Looks right so far..

Too long. Did you try substituting this back in both equations as a check?
are you sure its wrong? i double checked the arithmetic and i found it correct. i also plugged the numbers back into both equations, and they are equal.
 
  • #4
Something seems off. What value are you using for g? The problem uses imperial units but in part b) you give the answer in metric.

The other catch is that in part a) you define the distance y=0 to be the bottom of the shaft. But clearly it must have traveled some distance first before reaching the speed of 8ft/s. So I would redefine y=0 to be the point at which the elevator reaches 8 ft/s.
 
  • #5
paisiello2 said:
Something seems off. What value are you using for g? The problem uses imperial units but in part b) you give the answer in metric.

The other catch is that in part a) you define the distance y=0 to be the bottom of the shaft. But clearly it must have traveled some distance first before reaching the speed of 8ft/s. So I would redefine y=0 to be the point at which the elevator reaches 8 ft/s.
yeah dang. i was using 9.8. my bad...
 
  • #6
My suggestion is to get into the habit of always, always plug in the units along with the numbers. You can catch lots of mistakes this way.
 

Related to One dimensional Kinematics help

1. What is one dimensional kinematics?

One dimensional kinematics is a branch of physics that deals with the motion of objects along a single straight line. It involves studying the position, velocity, and acceleration of an object without considering any other factors such as rotation or forces.

2. How is one dimensional kinematics different from two or three dimensional kinematics?

One dimensional kinematics only considers motion along a single straight line, while two or three dimensional kinematics involves motion in two or three dimensions, respectively. This means that objects in one dimensional motion only have a position, velocity, and acceleration along one axis, while objects in two or three dimensional motion have these characteristics along multiple axes.

3. What are the basic equations of one dimensional kinematics?

The basic equations of one dimensional kinematics are the equations of motion, which include:
- Displacement (Δx = xf - xi)
- Average velocity (vavg = Δx/Δt)
- Instantaneous velocity (v = dx/dt)
- Average acceleration (aavg = Δv/Δt)
- Instantaneous acceleration (a = dv/dt)
- Kinematic equations (vf = vi + at, Δx = vit + 1/2at^2, and vf^2 = vi^2 + 2aΔx)

4. How is one dimensional kinematics used in real life?

One dimensional kinematics is used in many real life applications, such as calculating the speed of a vehicle, determining the acceleration of an object in free fall, and predicting the trajectory of a projectile. It is also used in sports, such as calculating the velocity of a baseball or the acceleration of a runner.

5. What are some common misconceptions about one dimensional kinematics?

Some common misconceptions about one dimensional kinematics include thinking that acceleration always means an increase in speed, when it actually refers to any change in velocity; assuming that an object at rest has a velocity of zero, when it actually has a velocity of zero at that moment; and thinking that the direction of motion is always the same as the direction of acceleration, when it can be in the opposite direction depending on the circumstances.

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