The acceleration of a particle moving only on a horizontal xy plane is given by , where is in meters per second-squared and t is in seconds. At t = 0, the position vector locates the paticle, which then has the velocity vector . At t = 3.60 s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?
In fact you left all the quantities out !
acceleration of a particle moving only on a horizontal xy plane is given by ****** , where ************ is in meters per second-squared and t is in seconds. At t = 0, the position vector ************locates the paticle, which then has the velocity vector ******************.
To find the position vector of the particle at t = 3.60 s, we need to integrate the acceleration function twice with respect to time to obtain the position function. Let's go through the steps:
1. First, let's integrate the acceleration function with respect to time to find the velocity function:
v(t) = ∫a(t) dt
Since the acceleration function is given as = 6.0t - 4.0t², we integrate it to get the velocity function:
v(t) = ∫(6.0t - 4.0t²) dt
= 6.0 * ∫t dt - 4.0 * ∫t² dt
= 6.0 * (0.5t²) - 4.0 * (1/3)t³ + C₁
Simplifying the expression, we get:
v(t) = 3.0t² - (4/3)t³ + C₁
2. Next, we integrate the velocity function with respect to time to find the position function:
r(t) = ∫v(t) dt
Since we know the velocity vector at t = 0 is <2.0, 0, 0> m/s, we can substitute this into the position function to find the constant C₁:
r(0) = 2.0 = 3.0(0)² - (4/3)(0)³ + C₁
2.0 = C₁
Now we can integrate the velocity function to get the position function:
r(t) = ∫(3.0t² - (4/3)t³) dt
= 3.0 * ∫t² dt - (4/3) * ∫t³ dt
= 3.0 * (1/3)t³ - (4/3) * (1/4)t⁴ + C₁t + C₂
Substituting C₁ = 2.0, we get:
r(t) = (1/3)t³ - (1/3)t⁴ + 2.0t + C₂
3. Finally, we substitute t = 3.60 s into the position function to find the position vector at t = 3.60 s:
r(3.60) = (1/3)(3.60)³ - (1/3)(3.60)⁴ + 2.0(3.60) + C₂
After evaluating this expression, we can find the position vector of the particle at t = 3.60 s in unit-vector notation.
To find the angle between its direction of travel and the positive direction of the x-axis, we need to calculate the direction vector or velocity vector at t = 3.60 s. We differentiate the position function with respect to time to get the velocity function, and then evaluate it at t = 3.60 s. We can then use the dot product to find the angle between two vectors.
Let's go through these calculations to find the position vector and the angle between the particle's direction of travel and the positive x-axis.