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
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?
3 answers
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
Because when
a_y= dv_y/dt = (v_x)^2 + a_x*x
= 25 + 5(2.5) = 32.5
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
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