A point moves along the Ox axis according to x =
A point moves along the $Ox$ axis according to $x = t^{3}-2t^{2}+3t+4$ (cm, s). Find the velocity and the acceleration at $t = 2\ \text{s}$.
Explanation of Method
The first two derivatives of a position law.
Step 1: Velocity
\[ v(t) = \dot{x} = 3t^{2}-4t+3. \]
\[ v(2) = 3\cdot4-4\cdot2+3 = 12-8+3 = 7\ \text{cm/s}. \]
Step 2: Acceleration
\[ a(t) = \ddot{x} = 6t-4, \qquad a(2) = 12-4 = 8\ \text{cm/s}^{2}. \]
Step 3: Reading the signs
Both are positive, so at $t = 2$ the point is moving in the positive direction and speeding up. In fact the velocity never vanishes: the discriminant of $3t^{2}-4t+3$ is $16-36 = -20\lt 0$, so $v>0$ for all $t$ and the motion never reverses.
Step 4: Verification
\[\text{Minimum speed at } t = \tfrac23:\ v = 3\cdot\tfrac49-\tfrac83+3 = \tfrac53 \approx 1.667\ \text{cm/s} \;\;\checkmark\]
\[a = 0 \text{ at } t = \tfrac23 \text{ too, which is exactly where the speed is least} \;\;\checkmark\]
\[\text{Numerically: } \frac{x(2.001)-x(1.999)}{0.002} \approx 7.000, \quad \frac{v(2.001)-v(1.999)}{0.002} = 8.000 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,v(2) = 7\ \text{cm/s}, \qquad a(2) = 8\ \text{cm/s}^{2}\,}\]
Remark
Remark. The acceleration $6t-4$ changes sign at $t = \tfrac23$, which is the inflection point of the position graph. Before it the point decelerates, after it accelerates; since the velocity stays positive throughout, that instant is a minimum of speed rather than a reversal of direction.