Question

Question 2.14: An eletron moving along the x-axis has a position given by x=16te^(-t)m where t is in seconds. How far is the electron from the origin when it stops?

Answer 1

x = 16t e^-t m

v = $\frac{\partial x}{\partial t} = 16 e^{-t} -16t e^{-t}$

$a = \frac{\partial v}{\partial t} = -16 e^{-t} -16 e^{-t} + 16t e^{-t} = -32e^{-t} + 16 t e^{-t}$

Stopping position is when v = 0 => t=1s

a, b, d in graph

Position Vs time Stopping position (1 sec) Location when min velocity 4 10 Time (s) Velocity Vs Time 10 6 Time (s) minimum velocity (at 2 s) acceleration 10 4 E -10 -15 2-20 25 35 Time (s)

c) acceleration when electron stops (t=1) = $\frac{-32}{e} +\frac{16}{e} = \frac{-16}{e} m/s^{2}$ = 5.886 m/s²

e) minimum velocity of particle is when $\frac{\partial v}{\partial t} =0$ => t =2sec

minimum velocity = $16 e^{-2} - 16*2e^{-2} = \frac{-16}{e^{2}} = -2.1653 m/s$

f) location of particle when minimum velocity (t=2s) = $16*2*e^{-2} = 4.331 m$

g) The particles velocity keeps decreasing, comes to a stop at 1 s. Then the velocity becomes negative, becomes minimum at 2 sec, the start becoming less negative. With increasing time velocity approaches 0 and particle approaches its initial position. All this can be clearly seen in the graphs.