Kinematics - calculating acceleration

AI Thread Summary
Jack drives at a constant speed of 25 m/s while Jill accelerates from rest at 4.0 x 10^-3 m/s². To find how long it takes Jill to catch up, the equations of motion for both are set equal: 25t = (1/2)(4.0E-3)t². Solving this leads to a quadratic equation, which can be simplified to find that t = 12500 seconds or approximately 3.47 hours. The discussion highlights the importance of correctly applying kinematic equations and recognizing constant speed versus acceleration.
TheronSimon
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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!
 
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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
 
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