MHB Mechanics- General motion in a straight line.

AI Thread Summary
The discussion focuses on solving a mechanics problem involving motion in a straight line, specifically addressing parts (a), (b), and (c). For part (a), the correct approach involves solving the cubic equation t^3 - 140t + 575 = 0 and substituting the value of t into the velocity equation v = 0.3t^2 + 14. In part (b), the velocity must be set to zero to find the corresponding time, which is then used in the position equation x = -0.1t^3 + 14t. The participant initially made a calculation error by solving a quadratic instead of a cubic equation but later corrected it and successfully found all answers. The discussion highlights the importance of correctly identifying and solving the equations for motion.
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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