When Will Two Objects With Different Speeds and Accelerations Meet?

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Two objects start from the same point, with one moving 4 seconds after the other, and they have different initial speeds and accelerations. The problem requires determining the time at which they meet, with the correct equations for distance under constant acceleration being crucial. The initial equations provided by the user were incorrect, and the correct formula is s = vit + at²/2. After some guidance, the discussion emphasizes the importance of using proper symbols for speed and distance in physics problems. The final answer for when the objects meet is determined to be 28 seconds.
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Homework Statement


Two objects start from same point whit 4s one after another traveling same direction whit
speed01 =5 m/s ; a01=0.5 m/s2
and
speed02=2 m/s ; a 02=1 m/22

\tau=4s

Find at what time the objects meet?

Answer:28 s

Homework Equations


d01=S01*t01*(a01+t012)/2
d02=S02*t02*(a02+t022)/2
(I think!)

The Attempt at a Solution


t01=S01/a01
t02=S02/a02

After fiding the distance from equation's i get stuck
 
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Welcome to PF!

Hi sese! Welcome to PF! :smile:

(btw, we usually use v or u for speed, not s, and s for distance, not d :wink:)
sese said:

Homework Equations


d01=S01*t01*(a01+t012)/2
d02=S02*t02*(a02+t022)/2

No, those equations are wrong.

You need s = vit + at2/2.

See the PF Library on constant acceleration :wink:
 

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