A question about Rectilinear motion

AI Thread Summary
The discussion revolves around a physics problem involving a particle moving under uniform acceleration and covering equal distances in varying time intervals. The average speed of the particle for the total distance of 3S is correctly identified as 3S / (t1 + t2 + t3), corresponding to option C. However, there is confusion regarding option B, which is also claimed to be correct, as it involves expressing average speeds in terms of initial and final velocities. Through further analysis, it is clarified that the average speed during the second interval can be expressed as the average of the speeds from the first and third intervals, leading to the conclusion that option B can also be justified. Ultimately, the problem is resolved with a clearer understanding of the relationships between average speeds and acceleration.
Priyadarshi Raj
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Homework Statement


Q:
If a particle moving along a straight line under uniform acceleration covers successive equal distances ( S each) in time intervals t1, t2 and t3 respectively, then the expression for average speed of particle in covering the given distance of 3S is:

Options:
A) (S/t1) + (S/t2) + (S/t3)
B) (S/t1) - (S/t2) + (S/t3)
C) 3S / (t1 + t2 + t3)
D) (S/t1) + (S/t2) - (S/t3)

[You may assume the particle is speeding up for the entire journey]

Here the correct answer are options B and C.
I cannot see how B is correct, although I easily got the option C.

Homework Equations


Average speed = Total distance covered / Total time taken

The Attempt at a Solution


Avg speed = (S + S + S) / (t1 + t2 +t3) = 3S / (t1+t2+t3)
This gives option C.
But I cannot get option B anyhow.
 
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Priyadarshi Raj said:
Avg speed = (S + S + S) / (t1 + t2 +t3) = 3S / (t1+t2+t3)
This gives option C.
Good.

Priyadarshi Raj said:
But I cannot get option B anyhow.
Hint: Each term in B is an average speed. Express those averages in terms of initial and final speed.
 
Doc Al said:
Hint: Each term in B is an average speed. Express those averages in terms of initial and final speed.

Okay then, I believe this is the way:
v2 will be average of v1 and v3
If the avg speeds are v1 = (S/t1) , v2 = (S/t2), v3 = (S/t3), then
Distance traveled during t2 is given by
S = v1t2 + 0.5at22 and also a = (v3 - v1)/t2 [using Newton's laws of motion]
∴ S = v1t2 + 0.5(v3 - v1)t2
⇒ (S/t2) = v1 + 0.5v3 - 0.5v1
⇒ 2(S/t2) = v1 + v3
⇒ 2(S/t2) = (S/t1) + (S/t3)
⇒ (S/t2) = (S/t1) - (S/t2) + (S/t3) = v2
⇒ vavg = v2 = (S/t1) - (S/t2) + (S/t3)

Am I correct?
 
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Priyadarshi Raj said:
Okay then, I believe this is the way:
v2 will be average of v1 and v3
If the avg speeds are v1 = (S/t1) , v2 = (S/t2), v3 = (S/t3), then
I'm not sure I follow your reasoning.

Here's how I would look at it. For interval t1, let the initial and final speeds be v1 and v2. Express the average speed (S/t1) in terms of v1 and v2. Do the same for the other intervals and the total.
 
Doc Al said:
I'm not sure I follow your reasoning.

Here's how I would look at it. For interval t1, let the initial and final speeds be v1 and v2. Express the average speed (S/t1) in terms of v1 and v2. Do the same for the other intervals and the total.

Thank you for helping me.
The problem is solved.
 
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