Question

 Let 𝑓(𝑡) = 𝑒 −(𝑡−1) 2 describes the position of a particle at time 𝑡 ≥ 0. a) What is the initial displacement? b) Find the critical points of 𝑓(𝑡). c) Find the intervals of positive and negative velocities of the particle. d) Find the time where particle changes from acceleration to deceleration and vice versa. e) Find the maximum displacement of the particle. f) Sketch the graph of 𝑓(𝑡). [

Answer 1

$$f(t)=e^{-(t-1)^2}, t\geq0$$

a)

$$f(0)=e^{-(0-1)^2}=e^{-1}$$

The initial displacement is $e^{-1}$ units.

b)

$$f'(t)=-2(t-1)e^{-(t-1)^2}$$

$$f'(t)=0=>-2(t-1)e^{-(t-1)^2}=0=>t=1$$

Critical point: $t=1$

$$f(1)=e^{-(1-1)^2}=1$$

c)

$$v(t)=f'(t)=-2(t-1)e^{-(t-1)^2}, t\geq0$$

Positive velocity of the particle: $t\in[0, 1)$

Negative velocity of the particle: $t\in(1, \infin)$

d)

$$a(t)=f''(t)=v'(t)=(-2(t-1)e^{-(t-1)^2})'$$

$$=(-2+4(t-1)^2)e^{-(t-1)^2}=2(2t^2-4t+1)e^{-(t-1)^2}$$

$$a(t)=0=>2(2t^2-4t+1)e^{-(t-1)^2}=0$$

$$2t^2-4t+1=0, t\geq 0$$

$$t=\dfrac{2\pm\sqrt{2}}{2}=1\pm\dfrac{\sqrt{2}}{2}$$

If $0\leq t<1-\dfrac{\sqrt{2}}{2},a(t)>0$

If $1-\dfrac{\sqrt{2}}{2}< t<1+\dfrac{\sqrt{2}}{2},a(t)<0$

If $t>1+\dfrac{\sqrt{2}}{2},a(t)>0$

Particle changes from acceleration to deceleration at $t=1-\dfrac{\sqrt{2}}{2}.$

Particle changes from deceleration to acceleration at $t=1+\dfrac{\sqrt{2}}{2}.$

e) Critical point: $t=1$

$$f(1)=e^{-(1-1)^2}=1$$

The maximum displacement of the particle is $1$ unit.

f) Sketch the graph of 𝑓(𝑡).