The particle travels along the path defined by the parabola y=0.5(x^2). If the component of velocity along the x-axis is v=5t [m/s], determine the particle's distance from the origin and the magnitude of its acceleration when t=1s. The initial condition is t=0, x=0 and y=0.
Can someone please tell me tha answers and how to work it please?
Sorry a_y will be 37.5 as we know 25 + 12.5 = 37.5
The method use to get a_y is by applying chain rule for y= 0.5x^2 And you will get dy/dt= x(x-dot)
Then apply product rule uv’+vu’
U=x
V=x-dot
dv/dt=x(x-double dot) + (x-dot)(x-dot)
dv/dt=2.5(5) + (5t)^2
When t=1
dv/dt=37.5
To find the particle's distance from the origin (0,0) when t=1s, we need to find the coordinates of the particle at that time.
Given that the component of velocity along the x-axis is v=5t, we can integrate it with respect to t to find the position of the particle along the x-axis:
x = ∫(5t) dt
x = 2.5t^2 + Cx
Since the particle starts at x=0 when t=0, we can substitute these values in to find the constant of integration Cx:
0 = 2.5(0)^2 + Cx
0 = Cx
Therefore, the equation for the position of the particle along the x-axis becomes:
x = 2.5t^2
Now, to find the distance from the origin, we need to find the value of y when x=2.5t^2. We can substitute this value into the equation of the parabola y=0.5(x^2):
y = 0.5(2.5t^2)^2
y = 0.5(6.25t^4)
y = 3.125t^4
Thus, the coordinates of the particle at t=1s are (2.5, 3.125).
To find the magnitude of the particle's acceleration when t=1s, we need to find the equation for acceleration. Acceleration is the derivative of velocity with respect to time:
a = d/dt (5t)
a = 5 (since the derivative of t with respect to t is 1)
Since acceleration is constant (5) and in the positive direction, its magnitude is simply the value itself, which is 5 m/s^2.
According to the method used by drwls is correct.but for the a_y its should be 32.5
Because when
a_y= dv_y/dt = (v_x)^2 + a_x*x
= 25 + 5(2.5) = 32.5
When t = 1, dx/dt = 5t = 5 m/s
From v = dx/dt = 5t, and the starting conditions, you can conclude that
x = (5/2)t^2.
x = 5/2 = 2.5 when t = 1
y = 0.5*(2.5)^2 = 3.125 when t = 1
The distance from the origin in sqrt[x^2 + y^2]. At t=1, this is
D = sqrt[2.5^2 + 3.125^2]= 4.002
The acceleration components are
a_x = d^2x/dt^2 = dv_x/dt = 5 @t=1)
and
a_y = d^2y/dt^2 = dv_y/dt
v_y = dy/dt = dy/dx*dx/dt = x*v_x =
a_y= dv_y/dt = (v_x)^2 + a_x*x
= 25 + 5*1 = 30