A particle leaves its initial position x0 at time t=0 moving in the positive x direction with speed v0 but undergoing acceleration of magnitude (a) in the negative x direction. Find expressions for a) the time when it returns to the position x0 and b) its speed when it passes that point.
To answer these questions, we need to use the equations of motion. Let's go through the steps:
a) The time when the particle returns to the position x0:
First, let's find the equation of motion for the particle. Since the particle is undergoing acceleration in the negative x direction, we can write the equation of motion as:
x(t) = x0 + v0t + (1/2)at^2
Next, we want to find the time when the particle returns to the initial position x0. This means that x(t) should be equal to x0. Setting x(t) = x0, we have:
x0 + v0t + (1/2)at^2 = x0
Subtracting x0 from both sides, we get:
v0t + (1/2)at^2 = 0
Factoring out t, we have:
t(v0 + (1/2)at) = 0
Since time t cannot be zero, we can divide both sides by (v0 + (1/2)at):
t = 0 / (v0 + (1/2)at)
t = 0
So, the particle returns to the position x0 at t = 0.
b) The speed when the particle passes that point:
To find the speed, we need the expression for velocity. The velocity v(t) can be found by taking the derivative of the position equation with respect to time:
v(t) = v0 + at
Since we know the particle is passing through the point x0, we can substitute t = 0 into the velocity equation:
v(0) = v0 + a(0)
v(0) = v0
Therefore, the speed when the particle passes the point x0 is equal to the initial speed v0.