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

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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
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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!
 
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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.
 
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