Basic Kinematics: Solving for an Object's Motion with Constant Acceleration

  • Thread starter Thread starter themilkman
  • Start date Start date
  • Tags Tags
    Kinematics Lost
Click For Summary
SUMMARY

The discussion focuses on solving a kinematics problem involving an object in motion with a constant acceleration of 15 m/s². At time t = 0, the object has a position of 0 meters and an initial velocity of -80 m/s. To find the object's position at t = 10 seconds, users are directed to utilize the kinematic equation: \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). Applying this formula yields the correct position of -50 meters, confirming option a as the solution.

PREREQUISITES
  • Understanding of basic kinematics principles
  • Familiarity with kinematic equations
  • Knowledge of constant acceleration concepts
  • Ability to perform basic algebraic calculations
NEXT STEPS
  • Review the kinematic equation \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \)
  • Practice solving problems involving constant acceleration
  • Explore additional resources on introductory physics concepts
  • Investigate the implications of negative velocity in motion analysis
USEFUL FOR

This discussion is beneficial for students studying introductory physics, particularly those struggling with kinematics and motion problems involving constant acceleration.

themilkman
Messages
2
Reaction score
0

Homework Statement


An object is out in free space, in motion along a straight line, with a constant acceleration of 15 m/s2.
At time t = 0, the object is at x = 0 meters, with a velocity of -80 m/s. Where is the object at t = 10s ?

a. - 50 m

b. 20 m

c. - 4 m

d. 75 m

Homework Equations





The Attempt at a Solution


I'm not even sure how to start this problem. This class for some reason does not use calculus, and the teacher is terrible so I have no idea what to do. My best guess would be that 70m and the way I tried this was assuming since V= -80 m/s and constant accel - 15m/s2 that t1= -65m and then to add 15 m every second until t10 = 70. Only problem is this is not an answer.

What should I do?
 
Physics news on Phys.org
At the Introductory Physics page where you posted this, there is a sticky at the top with different formulas in there. The link is <https://www.physicsforums.com/showthread.php?t=110015> and the formula you need is in there. It's in the second post that Doc Al posted under "General".

You have all the information you need. Starting position, starting velocity, acceleration, and time.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
935
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K