Velocity of moving particle?

Question

The velocity v of a particle moving in the xy plane is given by v = (6.0t - 4.0t^2) i + 2.0 j, with v in meters per second and t (> 0) is in seconds

When does the speed equal 10 m/s?

I've been trying for 30min but haven't been able to find it.

Answer 1

Omg, these questions are out to haunt me.  Suprisingly, I had these types of questios asked in a math class *honors algebra II to be exact** and I totally bombed it.   I barely passed - a 70.   I even think the teacher gave a curve or something cuz it was sooo hard.  I mean we are only tenth graders and that type of stuff should be left too at least 11th graders.   Try googling to help u or just ask your teacher   - shes the ones that grades them so she must know how to do the work.....

Answer 2

I think you take the square root of the sum of the squares and then set it equal to 10 and solve for t.  something like 10 = sqrt((6t-4t^2)^2+2^2)

Answer 3

What you have here is vector notation. You must first calculate the magnitude of the velocity by using v = (v1² + v2²)^(1/2).

10 m/s = ((6.0t - 4.0t²)² + (2.0)²)^(1/2)

100 m²/s² = (6.0 t - 4.0t²)² + 4.0

100 m²/s² = 16t^4 - 48t^3 + 36t² + 4.0

To solve for t, you must factor the quartic function by means of long division. I don't know of any other way to do this.

Answer 4

I couldn't do it either. My abortive results are:

Given that x² + y² = R², we have

(6t−4t²)² + 4 = 100

6t − 4t² = √96 = √16√6 = 4√6

Divide by 4.

1.5t − t² = √6

Set as standard quadratic equation of the form at² + bt + c = 0.

−t² + 1.5t − √6 = 0

where

a = −1   b = 1.5    c = −√6

t = (−b ± √[b²−4ac]) ÷ 2a

t = (−1.5 ± √[2.25 − {4 × −1 × −√6}]) ÷ (−2) = a complex value; b² − 4ac is negative.