the velocity of a particle moving in the xy plane?

Question

the velocity of a particle moving in the xy plane is given by v = (6.0t - 4.0t^2)i + 8.0j, with v in meters per second and t (> 0) is in seconds.

When does the speed equal 10m/s? I can't figure this out, although I figured out that the velocity is never zero and the acceleration is zero at .75seconds

Answer 1

Looks like the guys before convoluted it a little too much. Why square out the Vx in the first place?

V^2 = Vx^2 + Vy^2
100 = Vx^2 + 64
36 = Vx^2
Vx = +- 6

6t - 4t^2 = +- 6
0 = 2t^2 - 3t +- 3

Using +3 gives you complex roots, however -3 gives:
t = (3 +- SQRT(33)) / 4
= -.686, and... 2.186 sec

Answer 2

You've been given an expression for the velocity which has x and y components. To get the magnitude of this vector, (the speed), do the following

speed = SQRT(vx^2 + vy^2)

vx = 6t -4t^2. vx^2 = 16t^4 -48t^3 + 36t

vy = 8. vy^2 = 64

speed = SQRT(16t^4 - 48t^3 + 36t^2 + 64)

When does the speed = 10 m/s

10 = SQRT(16t^4 - 48t^3 + 36t^2 + 64)

100 = 16t^4 - 48t^3 + 36t^2 + 64

0 = 16t^4 - 48t^3 + 36t^2 - 36

0 = 4t^4 - 12t^3 + 9t^2 - 9

It should be solvable from there, i hope.

Answer 3

I don't think this question really has a solution.......
The magnitude of the velocity is:
{[(6t-4t^2)^2]+(8^2)}^1/2
when this is equated to 10 m/s and solved for t, you get complex roots....

What dudara says is right...but you get complex solutions which means there is no solution....