Trying to find displacement using velocity and acceleration

Click For Summary
SUMMARY

The discussion centers on calculating the displacement of a train that decelerates from an initial velocity of 60.0 m/s to 30.0 m/s at a uniform rate of 1.25 m/s². The relevant equation used is v² = vo² + 2ad. The user attempted to isolate the variable 'd' but incorrectly manipulated the equation, resulting in a nonsensical negative distance. The correct approach requires careful algebraic manipulation to ensure proper units and signs are maintained throughout the calculation.

PREREQUISITES
  • Understanding of kinematic equations, specifically v² = vo² + 2ad
  • Basic algebraic manipulation skills
  • Knowledge of units of measurement in physics, particularly velocity and acceleration
  • Concept of displacement in the context of motion
NEXT STEPS
  • Review kinematic equations and their applications in physics
  • Practice algebraic manipulation of equations to isolate variables
  • Study the relationship between velocity, acceleration, and displacement
  • Explore examples of uniform acceleration problems in physics
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of motion problems involving displacement, velocity, and acceleration.

mrsconfused
Messages
1
Reaction score
0

Homework Statement


A train traveling at a velocity of 60.0 m/s is slowed down at a uniform rate of 1.25 m/s2. How far will it travel before its velocity is 1/2 its original value?


Homework Equations


v2 = vo2 + 2ad


The Attempt at a Solution


I first need to isolate my variable, so 'd'. In which case, I believe the equation would then look like 'd=vo2 [itex]\div[/itex] v2+2a. Filling in this equation with the numbers then looks like: d=60m/s2[itex]\div[/itex]30m/s2+2(-1.25m/s2). But when I calculate it all out I'm left with d=-0.5m/s2. This does not make any sense though; it's not possible for the train to travel a negative amount of distance. And also, isn't displacement measured in units of length and not time? I can't figure out how to get rid of the seconds. Needless to say, I am very confused. Thanks in advance.
 
Physics news on Phys.org
mrsconfused said:
I believe the equation would then look like 'd=vo2 [itex]\div[/itex] v2+2a.
I don't. Try that manipulation again. If you're still stuck, post your detailed working.
 

Similar threads

Motion of an Object: Displacement & Average Velocity
  • · Replies 8 ·
Replies
8
Views
2K
Solving Vector Diagrams: Finding Displacements & Velocities
  • · Replies 5 ·
Replies
5
Views
2K
Distance and displacement confusion
  • · Replies 27 ·
Replies
27
Views
3K
with this tension problem -- Mass on an accelerating cable
  • · Replies 3 ·
Replies
3
Views
2K
Solve Physics Q: Acceleration & Velocity of Object
  • · Replies 3 ·
Replies
3
Views
2K
Acceleration & max displacement from velocity equation
  • · Replies 9 ·
Replies
9
Views
5K
Kinematic Equations Homework Solutions
  • · Replies 8 ·
Replies
8
Views
2K
Kinematics Question -- Two motorbikes accelerating at each other....
  • · Replies 14 ·
Replies
14
Views
2K
Can a Tiger Outrun a Lion in a 100 Meter Dash?
  • · Replies 1 ·
Replies
1
Views
2K
2 Questions about motion in 1 direction
  • · Replies 2 ·
Replies
2
Views
9K