How to find acceleration from distance and time?

  • Thread starter Thread starter SagarPatil
  • Start date Start date
  • Tags Tags
    Acceleration Time
Click For Summary

Homework Help Overview

The problem involves calculating acceleration from given distance and time measurements, as well as applying Newton's second law. The context is a linear air track lab with specific values provided for distance, mass, weight force, and time.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the kinematic equation to find acceleration, while also exploring a second method involving final velocity. Some participants question the arithmetic in the calculations and suggest re-evaluating the final steps.

Discussion Status

Participants are actively discussing the arithmetic involved in the calculations. There is acknowledgment of errors in the original poster's computations, and some guidance is provided to correct these mistakes. Multiple interpretations of the calculations are being explored.

Contextual Notes

There is uncertainty regarding the accuracy of the time measurement provided, which the original poster suggests might be incorrect. The discussion also reflects on the implications of using different methods to arrive at acceleration.

SagarPatil
Messages
34
Reaction score
2

Homework Statement


The problem is asking to find acceleration from distance and time measurements. It is also asking to find acceleration from Newtons second law

This was a linear air track lab if you're wondering.

Whats given:
Distance = 81 cm
Total mass moved = 200g
Weight Force = 1.471N (150g)
Time = 0.003 secs (This might be wrong, but forget it. I just want to know how to calculate acceleration)

Homework Equations


S=ut+0.5*a*t^2

Where
S = distance
u= Initial velocity
a= acceleration
t=time

The Attempt at a Solution


So,

0.81m = (0)(0.003)+(0.5)(0.003)^2(a)
0.81m=4.5x10^-6a
a= 1.8x10^-7 m/s^2

I am using the formulas from this website
http://www.mathalino.com/reviewer/engineering-mechanics/motion-particle
 
Physics news on Phys.org
Here is my second attempt

where v = final velocity

v+u=2s/t
v=2s-u/t
v=2(0.81)-0/0.003
v=540 m/s^2

v-u/t
540-0/0.003

a= 18,000 m/s^2
 
You seem to be having some trouble with the arithmetic.
SagarPatil said:
0.81m=4.5x10^-6a
a= 1.8x10^-7 m/s^2
All was fine until the last step. try that again.
SagarPatil said:
v=540 m/s^2
You mean m/s.
SagarPatil said:
540-0/0.003
a= 18,000 m/s^2
You mean (540-0)/0.003.
Again, you made an arithmetic mistake at the last step.
 
Yea thanks I got 180,000 m/s on both

On the first one 0.81/4.5x10^-6
=180,000 m/s

second one

540-0/0.003
=180,000 m/s

So the steps I did are correct?
 
SagarPatil said:
Yea thanks I got 180,000 m/s on both

On the first one 0.81/4.5x10^-6
=180,000 m/s

second one

540-0/0.003
=180,000 m/s

So the steps I did are correct?
Yes.
 
  • Like
Likes   Reactions: SagarPatil

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
971
Replies
25
Views
3K
Replies
8
Views
6K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K