Displacement, velocity, acceleration

1. Sep 3, 2009

joemama69

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 couldnt 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

2. Sep 3, 2009

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.

3. Sep 9, 2009

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

4. Sep 9, 2009

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.

5. Sep 9, 2009

joemama69

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

6. Sep 9, 2009

rl.bhat

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

7. Sep 10, 2009

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

8. Sep 10, 2009

rl.bhat

Here v = Sqrt[ (10t-3)2+(2t)2]
What is the derivative of this?

9. Sep 12, 2009

joemama69

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

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

10. Sep 13, 2009

joemama69

11. Sep 13, 2009

rl.bhat

12. Sep 13, 2009

rl.bhat

I have edited this post.

13. Sep 13, 2009

joemama69

no problem, thanks for all the rest of ur help