Forces and Motion problem-solving question

A high-speed "smart-sled" is tested on a horizontal track for emergency braking systems. The sled, initially travelling at a constant velocity $u$, passes a sensor at $t = 0$. At this instant, the braking system is engaged. The velocity $v$ of the sled as a function of time $t$ is recorded by an onboard computer. For the first 2.00 seconds of braking, the velocity follows the relationship: $$v(t) = u - kt^2$$ where $k$ is a constant of the braking system. After $t = 2.00$ s, the sled switches to a constant deceleration $a_c$ until it comes to a complete rest. Data recorded: - Initial velocity $u = 45.0\text{ m s}^{-1}$ - Velocity at $t = 2.00\text{ s}$ is $v_2 = 33.0\text{ m s}^{-1}$ - Total stopping distance from $t = 0$ is $D = 110\text{ m}$ (a) Determine the value and units of the constant $k$. [2] (b) Calculate the distance travelled by the sled during the first 2.00 seconds of braking. [3] (c) Determine the magnitude of the constant deceleration $a_c$ applied after $t = 2.00$ s. [4] (d) A safety engineer suggests that if the initial velocity $u$ were increased by 10%, the "thinking distance" (the distance travelled at constant $u$ before the brakes engage) would become significant. If the total stopping distance must not exceed 150 m, calculate the maximum allowable reaction time $t_r$ for this higher speed, assuming the braking profile ($k$ and $a_c$) remains identical. [4]

(a) Using the given equation $v(t) = u - kt^2$ at $t = 2.00$ s: $33.0 = 45.0 - k(2.00)^2$ $4k = 45.0 - 33.0 = 12.0$ $k = 3.00$ Units: Since $v$ is in $\text{m s}^{-1}$ and $t^2$ is in $\text{s}^2$, $k$ must be in $\text{m s}^{-3}$. **$k = 3.00\text{ m s}^{-3}$** (b) Distance is the area under the $v-t$ graph. Since the acceleration is non-linear for $0 \le t \le 2$, we integrate: $s_1 = \int_{0}^{2} (u - kt^2) dt = [ut - \frac{1}{3}kt^3]_0^2$ $s_1 = (45.0 \times 2.00) - \frac{1}{3}(3.00)(2.00)^3$ $s_1 = 90.0 - 8.0 = 82.0\text{ m}$ **$s_1 = 82.0\text{ m}$** (c) Total distance $D = s_1 + s_2 = 110\text{ m}$. $s_2 = 110 - 82.0 = 28.0\text{ m}$. For the second phase ($t > 2$), the motion is constant deceleration from $v_2 = 33.0\text{ m s}^{-1}$ to $v = 0$. Using $v^2 = u^2 + 2as$: $0^2 = (33.0)^2 + 2(-a_c)(28.0)$ $56.0 a_c = 1089$ $a_c = 19.446...$ **$a_c = 19.4\text{ m s}^{-2}$** (d) New initial velocity $u' = 1.10 \times 45.0 = 49.5\text{ m s}^{-1}$. The braking profile is identical, so $k = 3.00$ and $a_c = 19.446$. New braking distance $s_{brake} = s_1' + s_2'$. $s_1' = \int_{0}^{2} (u' - kt^2) dt = (49.5 \times 2) - 8 = 91.0\text{ m}$. Velocity at $t=2$ is $v_2' = 49.5 - 3(2)^2 = 37.5\text{ m s}^{-1}$. $s_2' = \frac{v^2 - u^2}{2a} = \frac{0 - (37.5)^2}{2(-19.446)} = \frac{1406.25}{38.892} = 36.158\text{ m}$. Total braking distance $D' = 91.0 + 36.158 = 127.158\text{ m}$. Available thinking distance $s_{think} = 150 - 127.158 = 22.842\text{ m}$. $t_r = \frac{s_{think}}{u'} = \frac{22.842}{49.5} = 0.4614...$ **$t_r = 0.461\text{ s}$**

(a) 2 marks: 1 for numerical value 3.00, 1 for correct units $\text{m s}^{-3}$. (b) 3 marks: 1 for identifying integration or area under curve, 1 for correct integration of $kt^2$, 1 for final answer 82.0 m. (c) 4 marks: 1 for finding $s_2 = 28.0$ m, 1 for selecting $v^2 = u^2 + 2as$, 1 for substituting $v_2 = 33.0$ as the initial velocity for this phase, 1 for final answer 19.4 $\text{m s}^{-2}$ (allow 19.5 if rounding occurred earlier). (d) 4 marks: 1 for new $u' = 49.5$, 1 for calculating new $s_1'$ and $s_2'$ (or total braking distance 127.2 m), 1 for finding available thinking distance 22.8 m, 1 for final reaction time 0.461 s. Common errors: Using $v=u+at$ for the first 2 seconds (it is non-linear); forgetting to change the starting velocity for the second phase in part (d).

Hint 1 (mild nudge): For part (a), substitute the known values at $t=2$ into the given equation. For part (b), remember that displacement is the area under a velocity-time graph; if the graph is a curve, you need calculus. Hint 2 (partial direction): In part (c), the "initial" velocity for the constant deceleration phase is the velocity the sled had at the end of the first phase ($33.0\text{ m s}^{-1}$). Hint 3 (key insight): In part (d), the total stopping distance is the sum of thinking distance ($u' \times t_r$) and the two-stage braking distance. You must recalculate the braking distance because the higher starting speed changes the velocity at every point during the first 2 seconds.