Please help with an acceleration and distance problem

  • Thread starter Thread starter science_rules
  • Start date Start date
  • Tags Tags
    Acceleration
Click For Summary
SUMMARY

The discussion focuses on calculating the negative acceleration and distance traveled by a rocket sled moving at 632 miles per hour, which comes to a stop in 1.40 seconds. The correct negative acceleration is -202 meters/second squared, while the distance traveled during this deceleration is 119 meters. The participant initially miscalculated the acceleration and distance due to incorrect unit conversions and the use of inappropriate formulas. The correct equations to use are v = u + at for acceleration and v² = u² + 2as for distance.

PREREQUISITES
  • Understanding of basic kinematics equations
  • Familiarity with unit conversions, particularly miles per hour to meters per second
  • Knowledge of acceleration and distance formulas in physics
  • Ability to interpret and apply negative values in physical equations
NEXT STEPS
  • Study the kinematic equations in detail, focusing on their applications
  • Practice unit conversion techniques, especially between different speed metrics
  • Explore examples of negative acceleration scenarios in physics
  • Learn about the implications of sign conventions in physics calculations
USEFUL FOR

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

science_rules
Messages
157
Reaction score
2

Homework Statement


A rocket sled goes down a track at a speed of 632 miles/hr. The sled is then brought to a rest in 1.40 seconds. Determine the a) negative acceleration experienced on the sled and b) the distance traveled during the negative acceleration.


Homework Equations


I attempted to calculate the acceleration using the a = change in velocity divided by change in time. The answer in my book was -202 meters/second squared, which i didn't understand because when I tried converting my result to met/sec, it did not match this. it was too much: -661.4 met/sec. i also tried calculating the distance traveled using the distance formula: x= xi + vit + 1/2 at^2 This also didn't work because I realized when I tried converting my answer to met/seconds, it was still way too much. The book's answer was 119 meters. My answer was 1, 495.2 meters. I thought that I used the correct formulas but there must be something I am missing.



The Attempt at a Solution


See above.
 
Physics news on Phys.org
Be carefull with signs!
The equations are quite simple, v = u + at for the acceleration and v2 = u2 + 2as for the distance.
 
ok thanks! i think i was not aware of a few of the simpler equations to use when solving these problems.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K