Question 4 of 6: Question 4 (paper Question IV) — Relative Motion of Two Vehicles at an Overpass (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2017-May. Candidates were required to answer any 2 of 3 questions in Part A (Statics) and any 2 of 3 in Part B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1–3 = paper Part A, I–III; Questions 4–6 = paper Part B, IV–VI).
Check — figure reconstruction. Question 5's impact angle (18°) is on the opposite side of vertical from the 20° release angle, not the same side. Both readings are corroborated below by clean, self-consistent numerical results (e.g. member D–F resolves to an exact 2.5 m, and the opposite-side impact reading alone reproduces the rising-then-falling trajectory the figure draws for sphere B).
Question 4 (paper Question IV) — Relative Motion of Two Vehicles at an Overpass (Part B · Dynamics, equal value)
Given. At $t=0$: vehicle B is at the overpass centre (origin), travelling at a constant 55 mi/hr along a road making 20° with the +x axis (below it); vehicle A is 1000 ft from the centre along a second road making 60° with the +x axis (i.e. 30° from the +y axis), travelling at $v_{A0}=35$ mi/hr and accelerating at a constant $4\ \text{ft/s}^2$ toward the centre.
Given data
Quantity
Value
Initial separation of A from centre
1000 ft
$v_{A0}$
35 mi/hr $=51.33$ ft/s
$a_A$ (constant, toward centre)
4 ft/s²
$v_B$ (constant)
55 mi/hr $=80.67$ ft/s
Road A direction (from +x axis)
60°
Road B direction (from +x axis)
−20°
Find. The relative displacement $\mathbf{r}_{A/B}$, velocity $\mathbf{v}_{A/B}$ and acceleration $\mathbf{a}_{A/B}$ of A with respect to B, evaluated at the instant A crosses the centre of the overpass.
[Figure not reproduced: Figure 4 (reconstructed). Roads A and B crossing at the overpass centre; angles measured from the coordinate axes as printed on the exam. See the official exam paper.]
Approach. Treat each vehicle's motion as one-dimensional along its own straight road; use rectilinear kinematics to find the time for A to travel the 1000 ft to the centre, evaluate each vehicle's position/velocity/acceleration as a plane vector at that instant, then take the vector difference $(\ )_A-(\ )_B$.
Time for A to reach the centre. Along road A, $1000=v_{A0}t+\tfrac12a_At^2=51.33t+2t^2$. Solving the positive root: $$\boxed{t=12.95\ \text{s}}$$
Speed of A at that instant. $v_A=v_{A0}+a_At=51.33+4(12.95)=\boxed{103.13\ \text{ft/s}}$, directed along road A, $\mathbf{v}_A=103.13(\cos60^\circ,\sin60^\circ)=(51.56,\,89.30)\ \text{ft/s}$.
Vector kinematics for each vehicle at $t=12.95$ s. With $\mathbf{u}_A=(\cos60^\circ,\sin60^\circ)$, $\mathbf{u}_B=(\cos(-20^\circ),\sin(-20^\circ))$: $$\mathbf{r}_A=-1000\,\mathbf{u}_A+v_{A0}t\,\mathbf{u}_A+\tfrac12a_At^2\mathbf{u}_A\approx(0,0)\ \text{ft (confirms A is at the centre)}$$ $$\mathbf{r}_B=v_Bt\,\mathbf{u}_B=80.67(12.95)\mathbf{u}_B=(981.5,\,-357.2)\ \text{ft}, \qquad \mathbf{v}_B=80.67\,\mathbf{u}_B=(75.79,\,-27.59)\ \text{ft/s}, \qquad \mathbf{a}_B=0$$
Relative motion. $$\mathbf{r}_{A/B}=\mathbf{r}_A-\mathbf{r}_B=(-981.5,\,357.2)\ \text{ft},\quad |\mathbf{r}_{A/B}|=1044.5\ \text{ft}$$ $$\mathbf{v}_{A/B}=\mathbf{v}_A-\mathbf{v}_B=(-24.24,\,116.90)\ \text{ft/s},\quad |\mathbf{v}_{A/B}|=119.39\ \text{ft/s}$$ $$\mathbf{a}_{A/B}=\mathbf{a}_A-\mathbf{a}_B=\mathbf{a}_A=4(\cos60^\circ,\sin60^\circ)=(2.0,\,3.464)\ \text{ft/s}^2,\quad |\mathbf{a}_{A/B}|=\boxed{4.0\ \text{ft/s}^2}$$ (the relative acceleration equals A's own acceleration exactly, since B's velocity is constant.)
