Ranking the accelerations of three vehicles

Problem

Three vehicles start from rest and reach the following velocities:

• Car A: $0 \to 100\;\text{km/h}$ in $8{.}0\;\text{s}$;
• Truck B: $0 \to 60\;\text{km/h}$ in $12\;\text{s}$;
• Motorbike C: $0 \to 80\;\text{km/h}$ in $4{.}0\;\text{s}$.

Rank the three vehicles by increasing average acceleration.

Solution

Definition. The average acceleration is the change in velocity divided by the time taken, $a = \Delta v / \Delta t$. Since all start from rest, $\Delta v$ equals the final velocity. We must first convert the velocities to $\text{m/s}$ using $1\;\text{km/h} = 1/3{.}6\;\text{m/s}$, otherwise the comparison is skewed.

Car A.

$v_A = \frac{100}{3{.}6} \approx 27{.}8\;\text{m/s}, \qquad a_A = \frac{27{.}8}{8{.}0} \approx 3{.}5\;\text{m/s}^2$

Truck B.

$v_B = \frac{60}{3{.}6} \approx 16{.}7\;\text{m/s}, \qquad a_B = \frac{16{.}7}{12} \approx 1{.}4\;\text{m/s}^2$

Motorbike C.

$v_C = \frac{80}{3{.}6} \approx 22{.}2\;\text{m/s}, \qquad a_C = \frac{22{.}2}{4{.}0} \approx 5{.}6\;\text{m/s}^2$

Increasing order. Comparing the three values $a_B \approx 1{.}4 < a_A \approx 3{.}5 < a_C \approx 5{.}6\;\text{m/s}^2$:

$\ev{a_B < a_A < a_C}$

Remark. The ranking cannot be read off the final velocity alone, nor the time alone: motorbike C, despite not having the highest final velocity, has the greatest acceleration because it reaches a high velocity in the shortest time. Truck B, with a low final velocity and a long time, has the smallest acceleration.

Collegamenti

Argomenti: Kinematics Concetti: Acceleration Competenze: Ranking reasoning