How to Solve Multi Phase Kinematics Problems?

[canadiankid]

1. I have been given the total distance (100m) and the total time (9.780s) as well as a speed (11.7m/s) at t₁ and a final speed (10.5m/s) at the end of the given time (9.780s). The initial velocity is zero at t=0. Acceleration is constant. I am not asking for the answer, just an algebra based formula or series of formulas to lead me in the right direction (question was on a midterm two years ago), since i do not have d, a or t for t₁. I need to find:
a)What is the duration of the first acceleration phase? (t=0 to t₁)
b)What is the constant acceleration during the first phase?
c)What is the constant negative acceleration during the second phase? (t₁ to 9.780s)


Im not sure what any relevant equations are as I cannot find one that doesn't require at least two of a,d or t.


I calculated the average acceleration from t=0 to t=9.780s (a=20.4499m/s²) but that seems useless to me. I drew the acceleration, velocity (labelled this one) and position graphs to re-enforce the broader concept.

Thanks in advance for the help!

 

[PeterO]

1. I have been given the total distance (100m) and the total time (9.780s) as well as a speed (11.7m/s) at t₁ and a final speed (10.5m/s) at the end of the given time (9.780s). The initial velocity is zero at t=0. Acceleration is constant. I am not asking for the answer, just an algebra based formula or series of formulas to lead me in the right direction (question was on a midterm two years ago), since i do not have d, a or t for t₁. I need to find:
a)What is the duration of the first acceleration phase? (t=0 to t₁)
b)What is the constant acceleration during the first phase?
c)What is the constant negative acceleration during the second phase? (t₁ to 9.780s)Im not sure what any relevant equations are as I cannot find one that doesn't require at least two of a,d or t.I calculated the average acceleration from t=0 to t=9.780s (a=20.4499m/s²) but that seems useless to me. I drew the acceleration, velocity (labelled this one) and position graphs to re-enforce the broader concept.

Thanks in advance for the help!

You need to make use of the formula s = t * (u+v)/2
Here
s = displacement (perhaps what you called d?)
u = initial velocity
v = final velocity
t = time

Translated to English that formula reads displacement equals time by average velocity.

For the first phase u = 0, v = 11.7, t = t₁
For the second phase u = 11.7, v = 10.5, t = 9.78 - t₁

The total of those two displacements = 100

This will enable you to solve for t₁, and once you have that it should all be easy

 

[canadiankid]

Ah yes it is a system of equations... I always over think those. Thank you very much!

And yes I did you d for displacement, however you are more correct as d would represent distance, not displacement.