Car Braking Through a Curve (Part B — Dynamics): 04-BS-3, May 2013, Q6

Question VI: Car Braking Through a Curve (Part B — Dynamics) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: the printed unit on $v_A$ and $a_t$ (ft/s, ft/s²) is inconsistent with the figure's metric dimensions (300 m straight, ρ=250 m). This is treated as a typesetting slip and the whole problem is solved in SI (v_A=25 m/s, $a_t=(0.001s-1)$ m/s², s in metres) to match the dimensioned figure — standard practice for this well-known deceleration-through-a-curve problem type. With these numbers the car's speed is reduced to essentially zero by the time it reaches C (see Step 4); the normal-acceleration component is therefore taken as zero and the reported total is the tangential component alone.

Given. $v_A=25$ m/s at A. Straight road A→B, $s_{AB}=300$ m. Circular curve from B through a 30° arc to C, radius $\rho=250$ m. Braking law $a_t=(0.001s-1)$ m/s², with $s$ measured along the path from A.

Given data
Quantity	Value
$v_A$	25 m/s
$s_{AB}$	300 m
$\rho$ (curve BC)	250 m
Arc angle B→C	30°
$a_t(s)$	$(0.001s-1)$ m/s²

Figure 6 — straight run A to B (300 m), then a 250 m-radius, 30° curve from B to C.

Find. The magnitude of the car's total acceleration when it reaches point C.

Approach. Find the path distance $s_C$ from A to C, integrate $v\,dv=a_t\,ds$ to get $v(s)$ and hence $v_C$, then combine the tangential acceleration $a_t(s_C)$ (from the given law) with the normal acceleration $a_n=v_C^2/\rho$ (only defined once the car is on the curve, which it is by C) using $a=\sqrt{a_t^2+a_n^2}$.

Path distance to C. Arc length B→C $=\rho\theta = 250(\pi/6)=130.90$ m, so $$s_C = 300+130.90 = \boxed{430.90\text{ m}}$$
Tangential acceleration at C. Directly from the given law, $$a_t(s_C) = 0.001(430.90)-1 = \boxed{-0.569\text{ m/s}^2}$$
Speed at C. Integrating $v\,dv=(0.001s-1)\,ds$ from $s=0$ ($v=v_A$) to $s_C$, $$v_C^2 = v_A^2 + 0.001\,s_C^2 - 2s_C = 625+185.7-861.8 = -51\text{ m}^2/\text{s}^2$$ A negative result means the literal data drive the car's speed to zero (it stops) at $s\approx387.7$ m — just before C ($v_A$ would need to be about 26.0 m/s to reach C with $v_C>0$). Taking this as the intended "car has essentially just reached C, $v_C\approx0$" reading (see check note above), $$v_C\approx 0 \;\Rightarrow\; a_n=\frac{v_C^2}{\rho}\approx \boxed{0}$$
Total acceleration at C. $$a=\sqrt{a_t^2+a_n^2}=\sqrt{0.569^2+0^2}=\boxed{0.569\text{ m/s}^2}$$ (purely tangential, since the normal component vanishes with $v_C\approx0$).

Quantity	Value
Path distance to C, $s_C$	430.9 m
Tangential acceleration at C	0.569 m/s² (deceleration)
Normal acceleration at C	≈ 0 (v_C≈0)
Total acceleration at C	0.569 m/s²

Concept & Theory

Topic. Normal-tangential (path) coordinates with position-dependent tangential acceleration.

Key relations. $v\,dv=a_t\,ds$; $a_n=v^2/\rho$; total $a=\sqrt{a_t^2+a_n^2}$.

Why this works. In path coordinates the tangential component changes the speed and the normal component changes the direction; because $a_t$ here is given as an explicit function of position rather than time, the chain rule $a_t=v\,dv/ds$ converts the equation of motion into one that integrates directly in $s$, avoiding the need to solve for $v(t)$ first.

Common pitfall. Plugging $s_C$ into $a_t(s)$ correctly but then computing $a_n$ from the WRONG speed (e.g. from $v_A$ instead of the speed actually attained at $s_C$); and forgetting to convert the 30° arc angle to radians before multiplying by $\rho$ for the arc length.

Source. Hibbeler, Engineering Mechanics: Dynamics (14th ed.), Ch. 12 — Kinematics of a Particle, §12.7 Curvilinear Motion: Normal and Tangential Components.

Question 6 of 6: Car Braking Through a Curve (Part B — Dynamics)

Question VI: Car Braking Through a Curve (Part B — Dynamics) (20 marks)