Displacement, velocity, acceleration

AI Thread Summary
The discussion revolves around a particle's motion in the x-y plane, requiring calculations for displacement, velocity, and acceleration at a specific time, as well as identifying when velocity and acceleration are perpendicular. At t = 2, the magnitude of displacement is calculated to be 15.23, with velocity and acceleration magnitudes of 17.46 and 10.20, respectively. The method for finding when velocity and acceleration are at right angles involves taking the derivative of velocity and setting it to zero, leading to confusion about whether a solution exists. The average velocity between t = 0 and 1 is confirmed to be 10.20 m/s, which coincidentally matches the magnitude of acceleration. The thread concludes with clarifications on the derivative calculations and the approach to solving the problem.
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Homework Statement



consider a particle moving in the x-y plane as follows

x = 5t^2 - 3t
y = 2 + t^2

a) find the mag of the displacement, vel, and accel at t = 2
b)find the time when the vel and the accel are at right angles to each other
c) find the average vel between t = 0 and 1

Homework Equations





The Attempt at a Solution



a... x(2) = 14, y(2) = 6, mag of displacement is sqrt(14^2 + 6^2) = 15.23

vx(2) = 10(2) - 3 = 17, vy(2) = 2(2) = 4, mad of disp = 17.46

ax(2) = 10, ay(2) = 2, mag of disp = 10.20

b... i tried to graph them on my calc and find the intersection but i couldn't find the intersection... is this an appropriate method

c... average veolicy of x from t = 0 to 1 is... (7-(-3))/(1-0) = 10, for y... (2-0)/(1-0) = 2
the mag of the average is sqrt(10^2 + 2^2) = 10.20 m/s
 
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b) When the velocity acceleration are perpendicular, there is no change in the velocity. Take the derivative of velocity and equate it to zero.
 
"b) When the velocity acceleration are perpendicular, there is no change in the velocity. Take the derivative of velocity and equate it to zero. "

im a bit confused. the derivative of the velocity is a constant. does that mean there is no answer.


also is part c correct. it is the same answer as the mag of acceleration. is coincidence or what
 
Find vx and vy and ax and ay.
v = ( vx^2 + vy^2)^1/2
Find dv/dt and equate to zero. Since a is not equal to zero, magnitude of the velocity may be constant and acceleration is perpendicular to v.
 
vx = 10t - 3; vy = 2t

v = \sqrt{(10t-3)^2 + (2t)^2}
v' = .5(104t^2 - 60t + 9)^(-.5)(208t-60) = 0
i set just the top equeat to zero, 208t - 60 = 0, t = .29
 
joemama69 said:
vx = 10t - 3; vy = 2t

v = \sqrt{(10t-3)^2 + (2t)^2}
v' = .5(104t^2 - 60t + 9)^(-.5)(208t-60) = 0
i set just the top equeat to zero, 208t - 60 = 0, t = .29
Your v' is wrong.
It should be
v' = 0.5*{(10t-3)^2 + (2t)^2}^-1/2*{2(10t - 3) + 4t}
Now equate it to zero. Since the denominator cannot be zero, 2(10t - 3) + 4t = 0
 
"Now equate it to zero. Since the denominator cannot be zero, 2(10t - 3) + 4t = 0 "

if x = (10t-3)2+(2t)2, then dx = 2(10t-3)(10) + 2(2t)(2) = 200t - 60 + 8t = 208t-60
 
joemama69 said:
"Now equate it to zero. Since the denominator cannot be zero, 2(10t - 3) + 4t = 0 "

if x = (10t-3)2+(2t)2, then dx = 2(10t-3)(10) + 2(2t)(2) = 200t - 60 + 8t = 208t-60
Here v = Sqrt[ (10t-3)2+(2t)2]
What is the derivative of this?
 
but if u work out all the math inside the sqrt, you get

sqrt(104t^2 - 60t + 9)
 
  • #12
rl.bhat said:
It should be
v' = 0.5*{(10t-3)^2 + (2t)^2}^-1/2*{20(10t - 3) + 4t}
Now equate it to zero. Since the denominator cannot be zero, 20(10t - 3) + 4t = 0
I have edited this post.
 
  • #13
no problem, thanks for all the rest of ur help
 
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