A kinematics question refering to a delay time of vehicle with constant velocity before decelerating, what is minimum time?

Question

A motorist is traveling at a constant speed $v_0$ and is approaching a traffic light. He wishes to stop the car at the light. However, his reaction time, that is, the time it takes him to put his foot on the brakes, is $\tau$ and the maximum deceleration of the braked car is $a$. What is the minimum distance from the light in which he can bring his car to a stop?

My workings out are:

The distance covered in the delay period is integral of $v_0$ with respect to $\tau$, giving $s=\tau v_0$,

But the second part I don't understand, my logic is to simply integrate deceleration twice to give:

$\alpha=-a$, $v=-at$, $s=-\frac{1}{2}at^2$

Giving a final answer of $s=-\frac{1}{2}at^2 + \tau v_0$

However, the answer given is $s=-\frac{1}{2}z\left(\frac{v_0^2}{a}\right)+\tau v_0$

What have I missed here?

Answer 1

There are two things you must modify in your solution to arrive at the given solution:

1. It is good practice to not apply signs to variables; instead, use a negative when evaluating. (Some people may disagree with this, but textbooks will almost always use this rule.) Thus, your initial equation should just be $\alpha = a$.

2. Indefinite integrals must always include a constant, i.e. if $F(x)$ is an antiderivative of $f(x)$, then $\int{f(x)} = F(x) + C$.

When you integrate your initial equation $\alpha = a \,$, you should get $v = at + C \,$, and using the initial value yields $C = v_0 \,$, so the velocity equation should be

$$v = v_0 + at$$

Similarly, when you integrate the velocity equation, you should get $s = s_0 + v_ot + \dfrac{1}{2} at^2 \,$; since you're only concerned with displacement and not position, we can assume $s_0 = 0$ and get

$$s = v_0t + \dfrac{1}{2} at^2$$

Since the car comes to a stop, we can use the velocity equation to solve for $t$:

$$0 = v_0 + at \\ t = \dfrac{-v_0}{a} $$

Finally, substituting the above into the equation for $s$, we get

$$s = v_0 \left( \dfrac{-v_0}{a} \right) + \dfrac{1}{2} a\left( \dfrac{-v_0}{a} \right)^2 \\ \, \\ s = \dfrac{-v_0^2}{a} + \dfrac{1}{2} \left( \dfrac{v_0^2}{a} \right) \\ \, \\ s = -\dfrac{1}{2} \left( \dfrac{v_0^2}{a} \right) \\ $$

Then you can just tack on the $\tau v_0$ term to get the given solution.

Answer 2

His velocity starts reducing (initial velocity $v_0$) after brake is applied and comes to halt at time $t_s$ (say).

Change in velocity $ = 0 - v_0 = at_s$.

Or $ \, t_s = \frac{-v_0}{a}$ (where a is deceleration and hence negative).

Distance traveled during braking period $ \displaystyle = \int_0^{t_s} v \, dt = \int_0^{-v_0/a} - a \, t \, dt = -\frac{1}{2} a[t^2]_0^{-v_0/a} = -\frac{v_0^2}{2a}$

So total distance traveled $= Tv_0 - \frac{v_0^2}{2a}$