To find displacement at \( t = 1 \), integrate the velocity function:

\[ s(t) = \int v(t) \, dt = \int \left( -t^3 + \frac{7}{2}t^2 – 2t + 6 \right) dt \]

\[ s(t) = -\frac{1}{4}t^4 + \frac{7}{6}t^3 – t^2 + 6t + C \]

Since the object is at the origin at \( t = 0 \), \( C = 0 \).

Evaluate at \( t = 1 \):

\[ s(1) = -\frac{1}{4}(1)^4 + \frac{7}{6}(1)^3 – (1)^2 + 6(1) \]

\[ = -\frac{1}{4} + \frac{7}{6} – 1 + 6 = \frac{-3 + 14 – 12 + 72}{12} = \frac{71}{12} \text{ m} \]

Answer: Displacement at \( t = 1 \) is \( \boxed{\dfrac{71}{12}} \) meters

Acceleration is the derivative of velocity:

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}\left( -t^3 + \frac{7}{2}t^2 – 2t + 6 \right) \]

\[ a(t) = -3t^2 + 7t – 2 \]

Answer: Acceleration is \( \boxed{-3t^2 + 7t – 2} \) \( ms^{-2} \)

First find when the object comes to rest (\( v(t) = 0 \)):

\[ -t^3 + \frac{7}{2}t^2 – 2t + 6 = 0 \]

Solving gives \( t = 1.5 \) seconds (exact value may be \( \frac{3}{2} \))

To find maximum speed before \( t = 1.5 \), find when acceleration is zero:

\[ -3t^2 + 7t – 2 = 0 \]

\[ (3t – 1)(t – 2) = 0 \Rightarrow t = \frac{1}{3}, 2 \]

Only \( t = \frac{1}{3} \) is before 1.5 seconds.

Evaluate velocity at critical points and endpoints:

\[ v(0) = 6 \, ms^{-1} \]

\[ v\left(\frac{1}{3}\right) \approx 8 \, ms^{-1} \]

\[ v(1.5) = 0 \, ms^{-1} \]

Answer: Greatest speed before coming to rest is \( \boxed{8} \) \( ms^{-1} \)

Check all critical points and endpoints in \( 0 \leq t \leq 4 \):

\[ v(0) = 6 \, ms^{-1} \]

\[ v\left(\frac{1}{3}\right) \approx 8 \, ms^{-1} \]

\[ v(2) \approx 8 \, ms^{-1} \]

\[ v(4) = -64 + 56 – 8 + 6 = -10 \, ms^{-1} \]

The greatest magnitude of velocity is 10 \( ms^{-1} \).

Answer: Greatest speed is \( \boxed{10} \) \( ms^{-1} \)

Speed increases when velocity and acceleration have the same sign:

From analysis of \( v(t) \) and \( a(t) \), speed increases on:

1. \( \frac{1}{3} \leq t \leq 2 \) (both positive)

2. \( t \geq k \) (both negative)

The distance is the integral of absolute velocity over these intervals:

\[ \text{Distance} = \int_{\frac{1}{3}}^{2} v(t) \, dt + \int_{k}^{4} |v(t)| \, dt \]

Where \( k \) is when the object first comes to rest.

Answer: \( \boxed{\int_{\frac{1}{3}}^{2} v(t) \, dt + \int_{k}^{4} |v(t)| \, dt} \)