Find the total time taken and acceleration in the given problem-Kinematics

  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Acceleration Time
Click For Summary
SUMMARY

The forum discussion focuses on solving a kinematics problem involving total time and acceleration. The user derived equations for acceleration, total area under the velocity-time graph, and total time taken. The calculations resulted in an acceleration of 3.33 m/s² and a total time of 33 seconds. The discussion also highlights a potential error in the acceleration calculation, suggesting it should be based on a different time value.

PREREQUISITES
  • Kinematics equations for motion analysis
  • Understanding of velocity-time graphs
  • Basic algebra for solving equations
  • Concept of acceleration and its calculation
NEXT STEPS
  • Study the derivation of kinematic equations in detail
  • Learn how to analyze velocity-time graphs
  • Practice solving problems involving total distance and acceleration
  • Explore common mistakes in kinematics calculations
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in mastering motion analysis problems.

chwala
Gold Member
Messages
2,828
Reaction score
420
Homework Statement
Kindly see attached; interest only on parts;

a, c and d
Relevant Equations
Kinematics equations
This is the question; I made some math error...then i just realised this is an easy problem...anyway, i know you guys may have an alternative approach to this; kindly share...

1675764481515.png
For part (a) i have;

##a=\dfrac{10}{t_1}## and ##2a=\dfrac{20-10}{(t_1+t_2)-t_1}##

##⇒\dfrac{10}{t_1}=\dfrac{10}{2t_2}##

##t_1=2t_2##

For part (c); i have

##A_{total}= A_1+A_2+A_3##

where

##A_1=\dfrac{1}{2} × t_1 × 10##

##A_2=\dfrac{1}{2} × t_2 × (20+10)##

##A_3= 24 × 20##

##555=10t_2+15t_2+480##

##75=25t_2##

##t_2=3## seconds

##t_{total}=6+3+24=33##seconds

For part (d),

##a=\dfrac{10}{3}=3\frac{1}{3} m/s^2##

Cheers! Bingo!
 
Physics news on Phys.org
chwala said:
Homework Statement:: Kindly see attached; interest only on parts;

a, c and d
Relevant Equations:: Kinematics equations

This is the question; I made some math error...then i just realised this is an easy problem...anyway, i know you guys may have an alternative approach to this; kindly share...

View attachment 321883For part (a) i have;

##a=\dfrac{10}{t_1}## and ##2a=\dfrac{20-10}{(t_1+t_2)-t_1}##

##⇒\dfrac{10}{t_1}=\dfrac{10}{2t_2}##

##t_1=2t_2##

For part (c); i have

##A_{total}= A_1+A_2+A_3##

where

##A_1=\dfrac{1}{2} × t_1 × 10##

##A_2=\dfrac{1}{2} × t_2 × (20+10)##

##A_3= 24 × 20##

##555=10t_2+15t_2+480##

##75=25t_2##

##t_2=3## seconds

##t_{total}=6+3+24=33##seconds

For part (d),

##a=\dfrac{10}{3}=3\frac{1}{3} m/s^2##

Cheers! Bingo!
Yes, there does appear to be some math error.

I see that you skipped part (b), but the title of the thread,

"Solve the given problem involving the velocity-time graph",

does promise a graph.
 
SammyS said:
Yes, there does appear to be some math error.

I see that you skipped part (b), but the title of the thread,

"Solve the given problem involving the velocity-time graph",

does promise a graph.
part (b) is fine with me...i was interested on the highlighted part. I amended the thread title...

Cheers @SammyS
 
chwala said:
part (b) is fine with me...i was interested on the highlighted part. I amended the thread title...

Cheers @SammyS
I agree with a and c.
But I think in part d you've made a mistake.
You said a=10/t1 so a should be equal to 10/6.
 
  • Like
Likes   Reactions: chwala

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
25
Views
2K
Replies
23
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
38
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
10K
Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
1K
Replies
3
Views
2K