A proton moves along the x axis according to the equation x = 50t + 10t2, where x is in meters and t is in seconds. For the time range, -8 s ≤ t ≤ 4s, answer the questions below. Show all work and explain your answers.
a) Estimate the fathest distance this proton could move in the negative direction of the x axis. What is the velocity at the moment?
b) In this time range, find out when the proton slows down.
c) When it has a positive position on the axis, what is the minimum speed of the proton? In which direction is the particle moving at the time(s)?
a) The proton will be at the fathest distance in the negative x axis when the time derivative of x is zero, or the velocity is zero
velocity is zero when
The farthest distance on the negative x axis is
b) The proton slows down when the velocity is zero
that is at t = -2.5 seconds
c) The proton switches from postive axis to negative axis at x=0
This happens at two times.
Hence
The above equation has two roots.
At t = - 5 s
the speed is
At t = 0 s the speed is
So the minimum speed of the proton at both these times is 50 m/s
At t = - 5 s, the velocity is negative. Hence the proton is moving in the negative direction
At t = 0 s, the velocity is negative. Hence the proton is moving in the positive direction
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