Calculating Time for Object B to Catch Object A in Constant Velocity Motion

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SUMMARY

The problem involves calculating the time it takes for Object B, traveling at 3.0 m/s, to catch up with Object A, which travels at 2.0 m/s. Object A has a 1.0-minute head start. The equations of motion are established as d_A = 2.0t and d_B = 3.0(t + 1.0). Setting these distances equal leads to the equation 2.0t = 3.0t + 3.0, which simplifies to t = 3.0 seconds. Therefore, Object B will catch Object A after 3.0 seconds from the moment it starts moving.

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Kandy
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Object A travels at a constant velocity af 2.0 m/s east, and object B travels in the same direction at a constant velocity of 3.0 m/s. If object B starts 1.0 min after object A, how long (time) will it take object B to catch object A?
 
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Kandy said:
Object A travels at a constant velocity af 2.0 m/s east, and object B travels in the same direction at a constant velocity of 3.0 m/s. If object B starts 1.0 min after object A, how long (time) will it take object B to catch object A?

You're dealing with constant velocity so you know [itex]d = vt[/itex] applies.

Object A

[tex]d_A = 2.0t[/tex]

Object B

[tex]d_B = 3.0(t + 1.0)[/tex]

Because object B is going to catch object A, the distance will be the same. Can you take it from here?
 
I don't quite get it, but this is what I tried:

dB = 3.0(t+1.0)
= 3.0t+3.0

dA = dB
2.0t = 3.0t+3.0
-1.0t = 3.0
t=-3.0

but the answer key shows 1.2 x 10^2 s (2.0 min not -3.0 min)
sorry, but i need more help.
 

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