Home›Irodov›Problem 1.3

Irodov 1.3Trickykinematicsaverage-velocityv-t-graph

Car with Trapezoidal Velocity Profile

Problem 1.3

A car starts moving rectilinearly, first with acceleration

w = 5.0 \text{ m/s}^2

(the initial velocity is zero), then uniformly, and finally decelerating at the same rate

w

, comes to a stop. The total time of motion equals

\tau = 25 \text{ s}

. The average velocity during that time is

\langle v \rangle = 72 \text{ km/h} = 20 \text{ m/s}

. How long does the car move uniformly?

Answer:

\Delta t = 15 \text{ s}

Solution Path

In this problem the key was recognizing that the v-t graph is a trapezoid whose area equals the total distance. The equal acceleration and deceleration made it symmetric (one unknown t0). The quadratic gave t0 = 5 s, so the car cruises for 15 s.

01Problem Restatement

1/6

A car starts from rest on a straight road. It speeds up at a constant rate, then cruises at a steady speed, then slows down at the same constant rate until it stops.Given:
-

w = 5.0 \text{ m/s}^2

(magnitude of acceleration and deceleration)
-

\tau = 25 \text{ s}

(total time of the entire trip)
-

\langle v \rangle = 72 \text{ km/h} = 20 \text{ m/s}

(average velocity over the whole trip)We need to find:

\Delta t

, the time the car spends moving at constant speed, in seconds.

Find

\Delta t

(uniform motion time in seconds)

02Physical Picture and Strategy

2/6

Imagine watching this car from the roadside. First it pulls away from a traffic light, getting faster and faster. Then it settles into a cruising speed. Then it sees a red light ahead and brakes steadily until it stops.The velocity-time graph for this motion is a trapezoid: a rising straight line (acceleration), a flat top (uniform speed), and a falling straight line (deceleration). The slope of the rising part equals

+w

and the slope of the falling part equals

-w

.Strategy: The area under a velocity-time graph equals total distance traveled. We also know that total distance equals average velocity times total time. By equating these two expressions for distance, we can solve for the unknown time.

Trapezoid

v

t

graph. Area under graph = total distance =

\langle v \rangle \cdot \tau

03Step 1: Time Structure (Symmetry)

3/6

Since the car speeds up and slows down at the same rate

w

, the time to accelerate from rest to top speed equals the time to decelerate from top speed back to rest. We call each of these

t_0

.Using the definition of total time:

\tau = t_0 + \Delta t + t_0 = 2t_0 + \Delta t

where

t_0

is the acceleration (or deceleration) time in seconds, and

\Delta t

is the uniform motion time in seconds.This means: the 25-second trip is split into three parts, with the first and last parts being equal in duration.Using the first equation of motion (constant acceleration from rest,

u = 0

v_{\max} = u + wt_0 = 0 + w \cdot t_0 = w \cdot t_0

where

v_{\max}

is the maximum (cruising) speed in m/s.This means: the top speed the car reaches depends on how long it accelerates. Longer acceleration time means higher cruising speed.

\tau = 2t_0 + \Delta t

\quad v_{\max} = w \cdot t_0

04Step 2: Area Under v-t Graph Equals DistanceKEY INSIGHT

4/6

The total distance the car travels equals the area under the trapezoid in the

v

t

graph.Using the principle that area under a velocity-time graph equals displacement:

\text{Total distance} = \text{Area of trapezoid}

The trapezoid has parallel sides

\Delta t

(top) and

\tau

(bottom), and height

v_{\max}

\text{Area} = \tfrac{1}{2}(\Delta t + \tau) \times v_{\max}

We also know from the definition of average velocity:

\text{Total distance} = \langle v \rangle \cdot \tau

Setting these equal and substituting

v_{\max} = w t_0

and

\Delta t = \tau - 2t_0

\langle v \rangle \cdot \tau = w \, t_0 (\tau - t_0)

This means: the average velocity, the acceleration, and the total time together pin down exactly how long the car spent accelerating.

\langle v \rangle = \dfrac{w \, t_0 (\tau - t_0)}{\tau}

05Step 3: Solve and Reject

5/6

Now we have one equation with one unknown (

t_0

). We plug in the known acceleration, total time, and average velocity to find how long the car spent speeding up.Substituting into

\langle v \rangle = w \, t_0 (\tau - t_0) / \tau

\langle v \rangle \cdot \tau = w \, t_0 (\tau - t_0)

20 \text{ m/s} \times 25 \text{ s} = 5.0 \text{ m/s}^2 \times t_0 \times (25 \text{ s} - t_0)

500 \text{ m} = 5 \, t_0 (25 - t_0) \text{ m}

Simplifying:

t_0^2 - 25 \, t_0 + 100 = 0

Using the quadratic formula:

t_0 = \frac{25 \pm \sqrt{625 - 400}}{2} = \frac{25 \pm 15}{2}

Two roots:

t_0 = 5 \text{ s}

t_0 = 20 \text{ s}

We reject

t_0 = 20 \text{ s}

because:

2 t_0 = 2 \times 20 = 40 \text{ s} > \tau = 25 \text{ s}

Physically, this would mean the car spent 40 seconds just accelerating and decelerating, but the entire trip only lasted 25 seconds. Impossible.So

t_0 = 5 \text{ s}

, and:

\Delta t = \tau - 2t_0 = 25 - 10 = 15 \text{ s}

This means: the car accelerates for 5 seconds, cruises for 15 seconds, and brakes for 5 seconds.

t_0 = 5 \text{ s}, \quad \Delta t = 15 \text{ s}

06Verification and Summary

6/6

Verification: Plugging

t_0 = 5 \text{ s}

back in:

v_{\max} = w \cdot t_0 = 5.0 \text{ m/s}^2 \times 5 \text{ s} = 25 \text{ m/s}

\text{Trapezoid area} = \tfrac{1}{2}(\Delta t + \tau) \times v_{\max} = \tfrac{1}{2}(15 + 25) \times 25 = 500 \text{ m}

\langle v \rangle = \frac{500 \text{ m}}{25 \text{ s}} = 20 \text{ m/s} \; \checkmark

Checks out.Summary: In this problem the key was recognizing that the velocity-time graph is a trapezoid whose area equals the total distance traveled. The equal acceleration and deceleration rates made the problem symmetric, reducing it to one unknown (

t_0

). Once we set up the area equation and solved the resulting quadratic, only one root was physically possible. The car moves uniformly for 15 seconds.

\boxed{\Delta t = 15 \text{ s}}

Want more practice?

Try more PYQs from this chapter →