A body of mass a kg moves in a straight line according
A body of mass $a$ kg moves in a straight line according to $s = 1+t+t^{2}$. Find its kinetic energy $\dfrac{mv^{2}}{2}$ five seconds after the start of motion.
Explanation of Method
One differentiation, one substitution.
Step 1: Velocity
\[ v(t) = \frac{ds}{dt} = 1+2t. \]
Step 2: Evaluate at $t = 5$
\[ v(5) = 1+10 = 11\ \text{m/s}. \]
Step 3: Kinetic energy
\[ E = \frac{mv^{2}}{2} = \frac{a\cdot11^{2}}{2} = \frac{121a}{2} = 60.5\,a. \]
Step 4: Verification
\[\text{The motion is uniformly accelerated: } w = \frac{dv}{dt} = 2\ \text{m/s}^{2}, \text{ constant}\]
\[\text{Work--energy check: } \frac{mv^{2}}{2}-\frac{mv_{0}^{2}}{2} = a\frac{121-1}{2} = 60a = m\,w\,\Delta s = a\cdot2\cdot30 \;\;\checkmark\]
\[\Delta s = s(5)-s(0) = 31-1 = 30\ \text{m} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,E = \dfrac{121a}{2} = 60.5\,a\ \text{J}\,}\]
Remark
Remark. The constant term $1$ in $s = 1+t+t^{2}$ is the initial position and plays no part: differentiation removes it, so only the initial velocity $v_{0} = 1$ and the acceleration $2$ matter. The verification above is the work–energy theorem of Problem 329 in concrete form.