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

Homework Help Overview

The discussion revolves around a kinematics problem involving the calculation of total time taken and acceleration. Participants are analyzing different parts of the problem, specifically parts (a), (c), and (d), while referencing relevant kinematic equations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants share their calculations for acceleration and total time, while others express concerns about potential math errors. There is a mention of a skipped part (b) and a request for alternative approaches.

Discussion Status

Participants are actively discussing their findings and questioning each other's calculations. There is acknowledgment of possible errors, particularly in part (d), but no consensus has been reached on the correctness of the approaches taken.

Contextual Notes

One participant notes that part (b) was skipped, and there is a reference to a velocity-time graph that is expected to be part of the problem context.

chwala
Gold Member
Messages
2,828
Reaction score
425
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
2K
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