The equations of motion of a body are x(t) = bt^2 , y(t) =…

The equations of motion of a body are $x(t) = bt^2$, $y(t) = ct^2$, with $b = 4\ \mathrm{m/s^2}$ and $c = 2\ \mathrm{m/s^2}$. Find: (a) the equation of the trajectory $y = y(x)$ and sketch it; (b) the speed $v(t)$ and acceleration $a(t)$.

Explanation of Method

Motion along a straight line given by x(t) and y(t).

Step 1: (a) Trajectory

From $x = bt^2$ we get $t^2 = x/b$; substituting into $y = ct^2$: $y = \frac{c}{b}\,x = \frac{2}{4}\,x$.

\[y = \tfrac12\,x \quad (x \ge 0)\]

The path is a straight line through the origin with slope $\tfrac12$. Because $x = bt^2 \ge 0$, only the half-line $x \ge 0$ is travelled.

$y\ (\mathrm{m})$ $x\ (\mathrm{m})$ $t=1$ $t=2$ $t=3\ \mathrm{s}$ $4$ $16$ $36$ $8$ $18$ $y = \tfrac12 x$

The body moves along the straight line $y = \tfrac12 x$, starting from rest at the origin. Positions after 1, 2 and 3 s get further apart each second: uniformly accelerated motion along a line.

Step 2: (b) Speed

\[ v_x = 2bt = 8t, \]

\[ v_y = 2ct = 4t, \]

\[ v = \sqrt{v_x^2 + v_y^2} = 2t\sqrt{b^2 + c^2} = 2t\sqrt{20} = 4\sqrt5\,t. \]

Step 3: (b) Acceleration

\[ a_x = 2b = 8, \]

\[ a_y = 2c = 4, \]

\[ a = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt5. \]

\[\begin{gathered} v(t) = 4\sqrt5\,t \approx 8.94\,t\ \mathrm{m/s} \\ a = 4\sqrt5 \approx 8.94\ \mathrm{m/s^2}\ \text{(constant)} \end{gathered}\]

Result

\[\boxed{\,y = \tfrac12\,x \quad (x \ge 0)\,}\]

\[\boxed{\,\begin{gathered} v(t) = 4\sqrt5\,t \approx 8.94\,t\ \mathrm{m/s} \\ a = 4\sqrt5 \approx 8.94\ \mathrm{m/s^2}\ \text{(constant)} \end{gathered}\,}\]

Remark

Note. The body starts from rest and moves along a straight line with constant acceleration $4\sqrt5\ \mathrm{m/s^2}$, so $v = at$ as for any start from rest. Both $\vec v$ and $\vec a$ point along the line $y = x/2$.