Irodov 1.3: Symmetric Ramps Around an Unknown Plateau

Problem Irodov 1.3

A car moves along a straight line in three stages. Starting from rest, it first accelerates uniformly at $w = 5.0$ m/s $^2$ ; it then travels at a constant velocity; finally it decelerates, at the same rate $w$ , until it comes to rest. The entire trip, from start to stop, takes a total time $\tau = 25$ s, and the average velocity over that whole time is $\langle v\rangle = 72$ km/h. Find how long the car spends moving at constant velocity.

The velocity-time graph: a rise from rest at slope w lasting t_a, a flat top at wt_a lasting Δ t, then a fall to zero at slope -w, within the total time τ. None of the three durations is given. — The velocity-time graph: a rise from rest at slope $w$ lasting $t_a$ , a flat top at $wt_a$ lasting $\Delta t$ , then a fall to zero at slope $-w$ , within the total time $\tau$ . None of the three durations is given.

Given

$w = 5.0$ m/s $^2$
$\tau = 25$ s
$\langle v\rangle = 72$ km/h

Find

The duration

\Delta t

of the uniform leg

Solution

Step 1The two ramps are equal in duration

The car speeds up from rest and slows back to rest at the same rate $w$ , so the velocity-time graph is a trapezoid: a rising ramp of duration $t_a$ , a flat top at the speed $v_{\max}=wt_a$ reached, and a falling ramp back to zero. Because the two ramps share the same slope magnitude and both start or end at $v=0$ , they take the same time $t_a$ each, so

\tau = 2t_a + \Delta t, \qquad\text{i.e.}\qquad t_a = \frac{\tau-\Delta t}{2}. \tag{1}

Step 2Total distance is the trapezoid’s area

The area under the graph (the total distance covered) splits into the two triangular ramps plus the rectangle between them:

s = \frac12 w t_a^2 + w t_a \Delta t + \frac12 w t_a^2 = w t_a(t_a+\Delta t). \tag{2}

Relation (1) gives $t_a=(\tau-\Delta t)/2$ ; adding $\Delta t$ to both sides simplifies the sum $t_a+\Delta t$ that appears in this factored area.

t_a+\Delta t = \frac{\tau-\Delta t}{2}+\Delta t = \frac{\tau+\Delta t}{2}.

Substituting both $t_a$ and this simplified sum into $s=wt_a(t_a+\Delta t)$ gives

s = w\cdot\frac{\tau-\Delta t}{2}\cdot\frac{\tau+\Delta t}{2},

which multiplies out to

s = \dfrac{w(\tau^2-\Delta t^2)}{4}

(3)

Step 3Bring in the average velocity and solve

The average velocity is the total distance divided by the total time, which turns the second given into a second expression for the same distance $s$ :

s = \langle v\rangle\,\tau . \tag{4}

Setting (4) equal to (3) gives one equation in the single unknown $\Delta t$ :

\langle v\rangle \tau = \frac{w(\tau^2-\Delta t^2)}{4}.

Multiplying by $4/w$ and isolating the square,

\frac{4\langle v\rangle\tau}{w} = \tau^2-\Delta t^2, \qquad \Delta t^2 = \tau^2 - \frac{4\langle v\rangle\tau}{w}.

Pulling $\tau^2$ out of the right-hand side prepares the square root:

\Delta t^2 = \tau^2\left(1-\frac{4\langle v\rangle}{w\tau}\right), \qquad \Delta t = \tau\sqrt{1-\frac{4\langle v\rangle}{w\tau}}.

Substituting $\tau=25$ s, $\langle v\rangle=72$ km/h $=20$ m/s and $w=5.0$ m/s $^2$ , and evaluating the radicand,

\Delta t = 25\sqrt{1-\frac{4\times20}{5\times25}}\ \text{s} = 25\sqrt{1-0.64}\ \text{s} = 25\times 0.60\ \text{s}.

\Delta t = \tau\sqrt{1-\dfrac{4\langle v\rangle}{w\tau}} = 15\text{ s}

(5)

The same graph with the timings now filled in: the plateau lasts Δ t = 15 s, so each ramp takes t_a = 5 s and the top speed reached is wt_a = 25 m/s. — The same graph with the timings now filled in: the plateau lasts $\Delta t = 15$ s, so each ramp takes $t_a = 5$ s and the top speed reached is $wt_a = 25$ m/s.

Check

With $\langle v\rangle = 72$ km/h $=20$ m/s, the ramp time comes out $t_a = (25-15)/2 = 5$ s and the peak speed $wt_a = 25$ m/s $=90$ km/h, comfortably above the 72 km/h average, as a trapezoid’s peak must exceed its own mean. If $\langle v\rangle$ were pushed up to $w\tau/4 = 31.25$ m/s the radicand would hit zero and $\Delta t\to0$ : the car would have no time to cruise at all, which is the correct limit of an all-ramp (triangular) profile.

Remark

The clean trapezoid split in Step 2 relies entirely on the two rates being equal in magnitude ( $w$ up and $w$ down); had the car decelerated at a different rate $w'\neq w$ , the two ramps would take different times and relation (1) would become two unknowns wide, needing a second equation before (3)’s tidy form could be reached at all.