Displacement, velocity, acceleration

[joemama69]

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

v_x(2) = 10(2) - 3 = 17, v_y(2) = 2(2) = 4, mad of disp = 17.46

a_x(2) = 10, a_y(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

 

[rl.bhat]

b) When the velocity acceleration are perpendicular, there is no change in the velocity. Take the derivative of velocity and equate it to zero.

 

[joemama69]

"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

 

[rl.bhat]

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.

 

[joemama69]

v_x = 10t - 3; v_y = 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

 

[rl.bhat]

v_x = 10t - 3; v_y = 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

 

[joemama69]

"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

 

[rl.bhat]

"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?

 

[joemama69]

but if u work out all the math inside the sqrt, you get

sqrt(104t^2 - 60t + 9)

 

[joemama69]

I posted the question under calc homewokr help

https://www.physicsforums.com/showthread.php?t=336929

 

[rl.bhat]

I posted the question under calc homewokr help

https://www.physicsforums.com/showthread.php?t=336929

Sorry. It is my fault. Your derivative is correct.

 

[rl.bhat]

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.

 

[joemama69]

no problem, thanks for all the rest of ur help