Answer:
The magnitude of the particle's velocity at t = 2 s is 46.2 m/s.
Explanation:
Hi there!
The velocity of the particle is defined as the change in its position over time. Mathematically, it is expressed as the derivative of the position function with respect to time. The x and y-component of the velocity vector (vx and vy respectively) can be calculated as follows:
vx = dx/dt
vy = dy/dt
Let´s find first vx:
x(t) = 5 · t + t³
vx(t) = dx/dt = t + 3 · t²
Now, let´s find vy:
y(t) = 3 · t² + t⁴
vy(t) = dy/dt = 6 · t + 4 · t³
Now, let´s find the value of vx and vy at t = 2 s:
vx(t) = t + 3 · t²
vx(2) = 2 + 3 · 2² = 14
vy(t) = 6 · t + 4 · t³
vy(2) = 6 · 2 + 4 · 2³ = 44
At t = 2 s, the velocity vector of the particle is the following:
v = (14, 44) m/s
The magnitude of this vector is calculated as follows:
[tex]|v| = \sqrt{vx^{2} + vy^{2}} \\|v| = \sqrt{(14 m/s)^{2} + (44 m/s)^{2}} \\|v| = 46.2 m/s[/tex]
The magnitude of the particle's velocity at t = 2 s is 46.2 m/s.
