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

A body moves along the $x$-axis according to $x(t) = kt^2$, where $k = 10\ \mathrm{m/s^2}$. Find its velocity at $t = 5\ \mathrm{s}$.

Explanation of Method

Velocity as the slope of x(t).

Step 1: Velocity as a derivative

\[ v(t) = \frac{dx}{dt} = 2kt = 20t\ \ (\mathrm{m/s}). \]

Step 2: At 5 s

\[ v(5) = 2 \cdot 10 \cdot 5. \]

\[v(5\ \mathrm{s}) = 100\ \mathrm{m/s}\]

Check without calculus

Comparing $x = kt^2$ with $x = \tfrac12 a t^2$ shows uniform acceleration from rest with $a = 2k = 20\ \mathrm{m/s^2}$. Then $v = at = 20 \cdot 5 = 100\ \mathrm{m/s}\ \checkmark$

$x\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $2$ $4$ $5$ $6$ $100$ $250$ $350$ $100\ \mathrm{m}$ $1\ \mathrm{s}$ $\text{slope} = v(5) = 100\ \mathrm{m/s}$ $x = 10t^2$

position x(t)
tangent at t = 5 s

The velocity at an instant is the slope of the position–time curve there. At $t = 5\ \mathrm{s}$ the tangent rises 100 m for every 1 s.

Result

\[\boxed{\,v(5\ \mathrm{s}) = 100\ \mathrm{m/s}\,}\]