Kinematics - calculating acceleration

Click For Summary

Homework Help Overview

The problem involves kinematics, specifically calculating the time it takes for Jill, who starts from rest and accelerates, to catch up to Jack, who is driving at a constant speed. The scenario is set along a straight pathway with defined initial conditions for both individuals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion for both Jack and Jill, with some attempting to set their positions equal to find the time of catch-up. There are varying interpretations of the equations and methods to solve for time, including quadratic equations and simplifications.

Discussion Status

The discussion includes attempts to derive the time using different approaches, with some participants questioning the correctness of earlier calculations. There is acknowledgment of mistakes and misunderstandings regarding the application of kinematic equations, but no consensus has been reached on a final method or solution.

Contextual Notes

Some participants express confusion about the setup and the implications of Jack's constant speed versus Jill's acceleration. There are indications of missing clarity on how to properly apply the equations of motion in this context.

TheronSimon
Messages
37
Reaction score
0

Homework Statement



Jack is driving with a pail of water along a straight pathway at a steady 25 m/s when he passes Jill who is parked in her minivan waiting for him. When Jack is beside Jill, she begins accelerating at the rate of 4.0 x 10-3 m/s2 in the same direction that Jack is driving. How long does it take Jill to catch up to Jack?

Homework Equations





The Attempt at a Solution



I'm superbly and completely stumped by this, I really have no idea how to even begin solving this. If anyone could help that would be most appreciated!
 
Physics news on Phys.org
If Jack is driving at a steady 25 m/s, then kinematics tell us that:

x(t)=25t

Now, Jill, initially at rest, starts accelerating at the rate of 4.0E-3 m/s2, which results in:

x(t)=(1/2)(4.0E-3)t2

We are admitting that they both start their movement at the origin of (Oxy). You can now solve it for the same position (same x).
 
so 25 = 2x10^-3 (t)^2
25= SQRT0.002
25= 0.0447
t= 25/0.0447
t= 559s
or t = 9.3 minutes
 
TheronSimon said:
so 25 = 2x10^-3 (t)^2

25t, not 25. You must solve a quadratic equation.
 
you lost me sorry :'(
 
Mathoholic! said:
x(t)=25t

Mathoholic! said:
x(t)=(1/2)(4.0E-3)t2

These are the two equations. You forgot take the t term in the first equation into account.
 
so how about...
25 t^2 = .002 m/s^2 t^2 divide both sides by t^2
25 = .002 m/s^2 divide both by 0.002^2
so 25/ 0.002
so if we cancel out the m/s we are left with a sec so an answer of time of 12500 seconds or 3.47 hours.
 
Nice work, that is correct.
 
TheronSimon said:
so how about...
25 t^2 = .002 m/s^2 t^2 divide both sides by t^2
25 = .002 m/s^2 divide both by 0.002^2
so 25/ 0.002
so if we cancel out the m/s we are left with a sec so an answer of time of 12500 seconds or 3.47 hours.

It's partiallly wrong, Jack's speed is constant (thus no acceleration!).

25t=(1/2)(4.0E-3)t2 → (2.0E-3)t2-25t=0

You can either use the quadratic formula or cancel the product like so:

t=0 \vee t=25/(2.0E-3)=12500s

You got the solution right but you used a wrong method as 25≠0.002.
 
  • #10
Wow whoops, totally didn't see that xD
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
18
Views
2K
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
12
Views
2K
Replies
2
Views
4K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K