Velocity of Particle Homework: Find Acc, Zero Accel, & Speed=10 m/s

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The velocity of a particle in the xy plane is defined by the equation v = (5.9t - 4.1t²)i + 8.7j. The acceleration at t = 3.7 seconds is calculated to be -24.44i + 0j m/s², and the acceleration is zero at approximately 0.7195 seconds. To find when the speed equals 10 m/s, the equation 10 = √((5.9t - 4.1t²)² + 8.7²) is established. The discussion reveals confusion regarding the setup of this equation and how to manipulate it to isolate t. Clarification is provided on squaring both sides to eliminate the square root and rearranging the terms for further solving.
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Homework Statement


The velocity v of a particle moving in the xy plane is given by v= (5.9 t - 4.1 t2)i + 8.7j, with v in meters per second and t (> 0) in seconds. (a) What is the acceleration when t = 3.7 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) does the speed equal 10 m/s?


Homework Equations





The Attempt at a Solution


Part A is -24.44 i + 0 j m/s2

Part B is .7195 seconds.

How do I find part C?
 
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if velocity is v=xi+yj

then |v| =speed = √(x2+y2)
 
That makes sense, but how do I set that equation up. I am confused with the "t" being in the and the "t^2"?
 
tjbateh said:
That makes sense, but how do I set that equation up. I am confused with the "t" being in the and the "t^2"?

the 'y' term is a constant, so you can square both sides of the equation and then make the 'x2' the subject and then take the square root of both sides, then solve. (I hope you understood what I meant)
 
No, I'm sorry, It's just not making sense to me. So (5.9t-4.1t^2)^2??
 
tjbateh said:
No, I'm sorry, It's just not making sense to me. So (5.9t-4.1t^2)^2??

ok we'd get


10=\sqrt{(5.9t-4.1t^2)^2 +(8.7)^2}


so if you square both sides you get rid of the square root sign. Then rearrange and make (5.9t-4.1t^2)2 the subject and take the square root of both sides now.
 
would I get sometime like 1.3= t(5.9-4.1t)??
 
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