Constant Acceleration - A Special Case — Question 2

Question 2

A braking formula has an end to its domain

In an idealized straight-track test, a cart moves east at $18\ \mathrm{m/s}$ when an obstacle is detected 80 m ahead. It maintains that velocity for a reaction delay of 1.20 s, then has constant acceleration $-3.00\ \mathrm{m/s^2}$ until it first stops. After stopping it remains stationary. Treat all values as exact model data.

You may use $v=v_i+aT$ and $\Delta x=v_iT+\tfrac12aT^2$ separately on each stage. The cart is a point; clearance means the obstacle coordinate minus the cart coordinate.

1. Find reaction distance, braking time, braking distance, and total detection-to-stop time and distance.

2. Does the model stop the cart before the obstacle? Find the remaining clearance.

3. With the same initial speed and braking acceleration, find the largest reaction delay that brings the cart to rest exactly at the obstacle. State the condition for strictly positive clearance.

4. A student uses the braking formula for 8 s after braking begins and obtains a negative velocity. Explain why that value is outside this model’s braking domain, and give the actual cart position and velocity then, measured from detection.

Question 2 - Solution

A braking formula has an end to its domain

(a) Use a separate elapsed time for each stage.

Reaction distance is $18(1.20)=\boxed{21.6\ \mathrm{m}}$. During braking, setting $v=18-3T=0$ gives $\boxed{T_b=6.00\ \mathrm{s}}$. The braking displacement is $$18(6)-\tfrac12(3)(6)^2=\boxed{54.0\ \mathrm{m}}.$$ No reversal occurs during either moving stage, so these displacements are also distances. Total time is $\boxed{7.20\ \mathrm{s}}$ and total distance is $\boxed{75.6\ \mathrm{m}}$.

(b) Compare stopping position with the obstacle position.

The cart stops at 75.6 m relative to detection, leaving $$\boxed{80-75.6=4.4\ \mathrm{m}}$$ of positive clearance. This is a conclusion for the exact stipulated model.

(c) Allocate the remaining distance to the delay.

$$C(T_r)=80-18T_r-54=26-18T_r\quad\text{(metres; }T_r\text{ in seconds)}.$$

The braking distance remains 54 m, leaving 26 m for reaction travel. Hence the delay that stops exactly at the obstacle is $$\boxed{T_r=26/18=13/9\ \mathrm{s}\approx1.444\ \mathrm{s}}.$$ Strictly positive clearance requires $0\leq T_r<13/9\ \mathrm{s}$. Equality gives zero clearance, not a positive margin.

(d) Stop using the braking equation when its premise ends.

The expression $18-3(8)=-6\ \mathrm{m/s}$ extrapolates the negative acceleration for 2 s after the specified first stop. That would describe a different motion that reverses direction. In this model the acceleration becomes zero once the cart is at rest. Eight seconds after braking begins, the actual state is still $$\boxed{x=75.6\ \mathrm{m}\text{ from detection},\qquad v=0\ \mathrm{m/s}}.$$ The corresponding detection-clock time is 9.20 s. Algebraic formulas must be used within the time interval where their assumptions hold.