Question

A proton moves along the x axis according to the equation x =
50t + 10t^{2}, 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)?

Answer #1

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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