A body moves along the x -axis according to x(t) = A + Bt^2…

A body moves along the $x$-axis according to $x(t) = A + Bt^2 + Ct^3$, where $A = -1\ \mathrm{m}$, $B = 3\ \mathrm{m/s^2}$, $C = -2\ \mathrm{m/s^3}$. Find: (a) its velocity and acceleration; (b) the time after which it stops; (c) its maximum velocity and when it is reached.

Explanation of Method

Non-uniform acceleration: stopping time and maximum speed.

Step 1: (a) Velocity and acceleration

\[ v(t) = \frac{dx}{dt} = 2Bt + 3Ct^2 = 6t - 6t^2\ \ (\mathrm{m/s}), \]

\[ a(t) = \frac{dv}{dt} = 2B + 6Ct = 6 - 12t\ \ (\mathrm{m/s^2}). \]

\[v = 6t - 6t^2, \qquad a = 6 - 12t\]

The acceleration changes with time, so the formulas for uniform acceleration do not apply here.

Step 2: (b) When it stops

\[ v = 6t(1 - t) = 0 \;\Longrightarrow\; t = 0 \ \text{(the start) or } t = 1\ \mathrm{s}. \]

\[t = 1\ \mathrm{s}\]

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0.5$ $1$ $1.5$ $v_{\max} = 1.5\ \mathrm{m/s}\ \ (a = 0)$ $\text{stops, } v = 0$ $\text{moves backward}$

Velocity–time graph of $v = 6t - 6t^2$. The speed rises, peaks at 1.5 m/s at $t = 0.5\ \mathrm{s}$ (where the slope, the acceleration, is zero), and falls to zero at $t = 1\ \mathrm{s}$. After that $v$ is negative: the body turns around.

Step 3: (c) Maximum velocity

The velocity is largest when it stops growing, i.e. when the acceleration is zero: $6 - 12t = 0 \;\Longrightarrow\; t = 0.5\ \mathrm{s}$.

\[ v(0.5) = 6 \cdot 0.5 - 6 \cdot 0.25 = 1.5\ \mathrm{m/s}. \]

\[v_{\max} = 1.5\ \mathrm{m/s} \ \text{at}\ t = 0.5\ \mathrm{s}\]

Result

\[\boxed{\,v = 6t - 6t^2, \qquad a = 6 - 12t\,}\]

\[\boxed{\,t = 1\ \mathrm{s}\,}\]

\[\boxed{\,v_{\max} = 1.5\ \mathrm{m/s} \ \text{at}\ t = 0.5\ \mathrm{s}\,}\]

Remark

Note. Between 0 and 1 s the body moves from $x(0) = -1\ \mathrm{m}$ to $x(1) = -1 + 3 - 2 = 0$, so it covers 1 m. At $t = 1\ \mathrm{s}$ it stops only for an instant: after that $v \lt 0$ and it moves back in the $-x$ direction.