Mechanics- General motion in a straight line.

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Discussion Overview

The discussion revolves around solving a mechanics problem involving general motion in a straight line, specifically focusing on cubic equations related to velocity and position. Participants are addressing parts (a), (b), and (c) of the problem, which require calculations based on given equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest solving the cubic equation $t^3 - 140t + 575 =0$ for part (a) and then using the value of t in the equation $v=0.3t^2 + 14$ to find the final answer.
  • For part (b), it is proposed to set $v=0$ to find the time when velocity is zero and substitute that time into the equation $x= -0.1t^3 +14t$.
  • One participant calculates t as 4.24s but expresses confusion when substituting this value into the velocity equation, leading to a different result than expected.
  • Another participant points out a calculation error, stating that t=4.24 does not satisfy the equation.
  • For part (c), it is suggested to solve the cubic equation $x =- 0.1t^3 + 14t$ with $x=40$.
  • One participant acknowledges a mistake in solving a quadratic equation instead of the cubic equation and expresses gratitude for the guidance received.

Areas of Agreement / Disagreement

There is no consensus on the correct value of t for part (a) as participants express differing results and calculations. Some participants identify errors in calculations, while others confirm the need to solve cubic equations for the various parts of the problem.

Contextual Notes

Participants mention specific equations and calculations but do not resolve the discrepancies in their results. The discussion reflects a reliance on solving cubic equations and the potential for calculation errors in the process.

Shah 72
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Iam not able to get the ans for q(a) which is 6.5 m/s
I don't understand how to calculate (b)
 
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For (a) part you have to solve a cubic equation $t^3 - 140t + 575 =0 $ and put the value of t in $v=0.3t^2 + 14$ to get final answer.
for (b) part put $v=0 $ to get t for which velocity is zero and put that t in $x= -0.1t^3 +14t$ now you have to solve only.
 
DaalChawal said:
For (a) part you have to solve a cubic equation $t^3 - 140t + 575 =0 $ and put the value of t in $v=0.3t^2 + 14$ to get final answer.
for (b) part put $v=0 $ to get t for which velocity is zero and put that t in $x= -0.1t^3 +14t$ now you have to solve only.
I get t= 4.24s
When I substitute this value in v I get -0.3×4.24^2+14. I don't get the ans 6.5 m/s
 
DaalChawal said:
For (a) part you have to solve a cubic equation $t^3 - 140t + 575 =0 $ and put the value of t in $v=0.3t^2 + 14$ to get final answer.
for (b) part put $v=0 $ to get t for which velocity is zero and put that t in $x= -0.1t^3 +14t$ now you have to solve only.
Can you also pls guide me how to calculate (c)
 
Shah 72 said:
I get t= 4.24s
When I substitute this value in v I get -0.3×4.24^2+14. I don't get the answer 6.5 m/s

You are making a calculation error, t=4.24 does not satisfy the equation.
For (c) part again you have to solve the cubic equation in t $x =- 0.1t^3 + 14t$ put $x=40$
 
DaalChawal said:
For (a) part you have to solve a cubic equation $t^3 - 140t + 575 =0 $ and put the value of t in $v=0.3t^2 + 14$ to get final answer.
for (b) part put $v=0 $ to get t for which velocity is zero and put that t in $x= -0.1t^3 +14t$ now you have to solve only.
Iam so so sorry. I did a mistake. I have to solve the cubic equation. I did the mistake of solving quadratic equation.
 
DaalChawal said:
You are making a calculation error, t=4.24 does not satisfy the equation.
For (c) part again you have to solve the cubic equation in t $x =- 0.1t^3 + 14t$ put $x=40$
I realized the mistake. Thank you!
 
DaalChawal said:
You are making a calculation error, t=4.24 does not satisfy the equation.
For (c) part again you have to solve the cubic equation in t $x =- 0.1t^3 + 14t$ put $x=40$
Thanks a lottttt! I got all the ans
 

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