(a)
The acceleration is the derivative of velocity. Use the calculator to evaluate this numerical derivative..
[tex]v'(t) \approx -2.118[/tex]
_________________
(b)
x(t) is the position function, which is the antiderivative of velocity.
Via the fundamental theorem of calculus
[tex]\int_0^3 v(t)\,dt = x(3) - x(0) \implies x(3) = x(0) + \int_0^3 v(t)\,dt \implies \\ \\
x(3) = -5 + \int_0^3 v(t)\,dt = -1.760[/tex]
_________________
(c)
[tex]\int_0^{3.5} v(t)\, dt \approx 2.844[/tex]
The above quantity represents the particle's displacement, change in position,
for 0 ≤ t ≤ 3.5.
[tex]\int_0^{3.5} |v(t)|\, dt \approx 3.737[/tex]
The above quantity represents the particle's distance traveled on 0 ≤ t ≤ 3.5.
_________________
(c)
[tex]x_2(t) = t^2 - t \implies x_2'(t) = 2t - 1[/tex]
[tex]x_2'(t) = v(t) \implies t \approx 1.571[/tex]
t ≈ 1.571
