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
AI Thread Summary
The discussion revolves around solving a kinematics problem, where the user initially made a math error but later identified the solution as straightforward. For part (a), they derived that the acceleration is related to time intervals, concluding that t1 equals twice t2. In part (c), they calculated the total time taken as 33 seconds and found t2 to be 3 seconds. However, there was a suggestion that a mistake was made in part (d) regarding the calculation of acceleration. Overall, the conversation highlights the importance of accuracy in kinematic equations and peer review in problem-solving.
chwala
Gold Member
Messages
2,827
Reaction score
415
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top