Acceleration Problem: Solving for Constant Acceleration with Kinematic Equations

  • Thread starter Thread starter nordqvist11
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The discussion revolves around understanding displacement and acceleration in kinematic equations, particularly in the context of a car slowing down from 23 m/s to rest over 85 m. The original poster expresses confusion about determining the direction of displacement and acceleration when not specified in the problem. They calculated the acceleration as -3.1 m/s², which is correct, as it indicates deceleration opposite to the initial velocity direction. Clarification is provided that the acceleration must be negative due to the car's slowing down. The conversation highlights the importance of understanding vector directions in kinematics.
nordqvist11
Messages
15
Reaction score
0

Homework Statement



I'd just like someone to see if I answered correctly. I'm having a bit of difficulty grasping the vagueness of the displacement in these questions I've been getting in my textbook. From what I gather the displacement is relative to the vector points, (the two x points in kinematic equations). However, how am I to know what direction the car is traveling relative to the x plane if it doesn't specify the point in a question? I don't know if you guys get me.. The acceleration I found in the question below, could be positive or negative, it doesn't specify enough. Or am I wrong?

A car slows down from 23 m/s to rest in a distance of 85 m. What was its acceleration, assumed constant?

Homework Equations



The kinematic equations for constant acceleration.

I attempted to solve it with the following:

finalV^2=initialV^2+2a(finalX-initialX)

The Attempt at a Solution



Here is an image of my attempted work. I got -3.1 m/s^2 as my answer.
Physics_problem_solution.jpg
 
Last edited:
Physics news on Phys.org
Good work by you! :smile:
 
Yes! Thanks
 
The acceleration I found in the question below, could be positive or negative, it doesn't specify enough. Or am I wrong?

A car slows down from 23 m/s to rest in a distance of 85 m. What was its acceleration, assumed constant?

Sorry, I did not read your question carefully enough.

V0x = +23m/s (assumedly in the +x direction)
x0 = 0
x = 85

Since the car is decelerating, the acceleration vector will be in the opposite direction of the velocity vector, hence the appropriateness of the "-" in front of 3.1m/s2
 
Very helpful, thank you.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top