A motor car moving at a speed of 72 km/h can not come to a stop in less than 3.0 s while for a truck this time

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Q. A motor car moving at a speed of 72 km/h can not come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a higway the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5 s. (Comment: This is to illustrate why vehicles carry the message on the rear side. "Keep safe Distance")

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Concept understanding — Uniform Acceleration Kinematics

Uniform Acceleration Kinematics: From Intuition to Equations

Imagine you're riding a bicycle down a straight, gentle slope. You don't pedal — you just let gravity pull you. At first you're moving slowly, but soon you're flying. Your speed increases by the same amount every second. That's the core idea: uniform acceleration means your velocity changes by a fixed amount in each equal interval of time.

If you gain 2 m/s every second, your acceleration is $2 \, \text{m/s}^2$ . The "per second per second" is the key — it's a rate of change of a rate of change.

The Precise Statement

Uniform (or constant) acceleration means the acceleration vector does not change with time. In one dimension (straight-line motion), this gives us five interconnected quantities:

$u$ — initial velocity (at $t = 0$ )
$v$ — final velocity (at time $t$ )
$a$ — constant acceleration
$t$ — time elapsed
$s$ — displacement (change in position)

These are linked by three equations, often called the equations of motion for constant acceleration:

v = u + at

Each equation omits one variable — the first omits $s$ , the second omits $v$ , the third omits $t$ . This is deliberate: you pick the equation that matches the quantities you know and the one you need.

Where Do These Come From?

Start with the definition of acceleration: $a = \frac{\Delta v}{\Delta t}$ . For constant $a$ , this becomes $a = \frac{v - u}{t}$ . Rearranging gives $v = u + at$ — the first equation.

Now, for constant acceleration, velocity changes linearly with time. The average velocity is simply the arithmetic mean of initial and final velocities: $\frac{u + v}{2}$ . Displacement is average velocity times time:

s = \frac{u + v}{2} \cdot t

Substitute $v = u + at$ into this:

s = \frac{u + (u + at)}{2} \cdot t = \frac{2u + at}{2} \cdot t = ut + \frac{1}{2}at^2

That's the second equation.

The third equation comes from eliminating $t$ . From $v = u + at$ , we get $t = \frac{v - u}{a}$ . Substitute into $s = \frac{u+v}{2} \cdot t$ :

s = \frac{u+v}{2} \cdot \frac{v-u}{a} = \frac{v^2 - u^2}{2a}

Multiply both sides by $2a$ : $v^2 - u^2 = 2as$ , or $v^2 = u^2 + 2as$ .

Note

These equations assume acceleration is constant. If acceleration changes, these formulas break — you'd need calculus.

A Worked Example

A car starts from rest and accelerates uniformly at $3 \, \text{m/s}^2$ for $5$ seconds. Find its final velocity and the distance travelled.

Given: $u = 0$ , $a = 3$ , $t = 5$

Final velocity: $v = u + at = 0 + 3 \times 5 = 15 \, \text{m/s}$

Distance: $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 3 \times 25 = 37.5 \, \text{m}$

You could also use $v^2 = u^2 + 2as$ to check: $15^2 = 0 + 2 \times 3 \times s$ gives $225 = 6s$ , so $s = 37.5 \, \text{m}$ . Consistent.

Common Pitfalls …

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Using uniform acceleration kinematics, the minimum safe distance is the sum of the distance covered during the human reaction time (at constant speed) plus the extra distance the car needs to stop compared to the truck — the answer is 1.25 m.

The core idea is simple: both vehicles start at the same speed (72 km/h), but the truck can stop in 5.0 s while the car can stop in 3.0 s. That means the car has a shorter stopping distance than the truck — so why is there any risk of collision? Because the driver of the car does not react instantly. There is a 0.5 s delay before the car even begins braking. During that half-second, the car continues at full speed, closing the gap. The question asks: how much initial separation is needed so that even after that reaction delay, the car can still stop without hitting the truck?

We treat both vehicles as undergoing uniform acceleration (constant deceleration) once braking begins. The truck brakes for 5.0 s, the car for 3.0 s. The car's reaction time is a period of uniform motion (no braking). The condition for no collision is that the car's total stopping distance (reaction distance + braking distance) is less than or equal to the truck's stopping distance plus the initial separation.

Let's convert the speed to m/s first, because the times are in seconds.

Convert speed

$72 \text{ km/h} = 72 \times \frac{1000}{3600} = 20 \text{ m/s}$ .
Find the deceleration of each vehicle

For uniform acceleration from initial speed $u$ to rest ( $v=0$ ) in time $t$ ,

$a = \frac{v-u}{t} = \frac{0-20}{t} = -\frac{20}{t}$ .

So:
- Car: $a_c = -\frac{20}{3} \approx -6.67 \text{ m/s}^2$
- Truck: $a_t = -\frac{20}{5} = -4 \text{ m/s}^2$
The negative sign just means deceleration.
Stopping distance for each vehicle

Using $s = ut + \frac{1}{2}at^2$ (or $v^2 = u^2 + 2as$ ):
- Truck: $s_t = 20 \times 5 + \frac{1}{2}(-4)(5)^2 = 100 - 50 = 50 \text{ m}$
- Car (braking only): $s_{c,\text{brake}} = 20 \times 3 + \frac{1}{2}\left(-\frac{20}{3}\right)(3)^2 = 60 - 30 = 30 \text{ m}$
Tip

Notice the car's braking distance (30 m) is less than the truck's (50 m). If both started braking at the same instant, the car would stop well before hitting the truck. The danger comes entirely from the reaction delay.
Distance covered during reaction time

The car travels at constant 20 m/s for 0.5 s before braking:

$s_{\text{reaction}} = 20 \times 0.5 = 10 \text{ m}$ .
Total distance the car travels from the moment the truck signals

$s_{c,\text{total}} = 10 + 30 = 40 \text{ m}$ .
The condition for no collision

Let $d$ be the initial separation (distance from car's front to truck's rear). We must track the actual gap as a function of time, not just compare the two totals.

Let $t=0$ be the moment the truck signals.
- Truck rear position (measured from its own start): $x_t(t) = 20t - 2t^2$ for $t \le 5$ . …

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Concept: The Safe Distance as the Area Under the Closing-Rate Graph

Method: Track ONE Quantity — the Rate the Gap is Shrinking — and Integrate It, Instead of Two Separate Position Functions

The stored answer builds full position functions $x_c(t)$ and $x_t(t)$ for the car and truck separately and subtracts them. This method instead defines a single quantity — the closing rate, $v_{\text{rel}}(t) = v_{\text{car}}(t)-v_{\text{truck}}(t)$ — sketches its graph directly, and reads the minimum safe distance off as the area under that one graph, up to the moment the closing rate returns to zero.

Setup

Both vehicles start at $72\ \text{km/h} = 20\ \text{m/s}$ . Truck decelerates uniformly to stop in $5.0\ \text{s}$ : $a_t = -20/5=-4\ \text{m/s}^2$ . Car decelerates uniformly to stop in $3.0\ \text{s}$ once it starts braking: $a_c = -20/3\ \text{m/s}^2$ ; but the car only starts braking after a $0.5\ \text{s}$ human reaction delay, during which it moves at constant $20\ \text{m/s}$ .

Steps

Write the closing rate over the reaction period, $0\le t\le0.5\ \text{s}$ (car not yet braking, truck already is):

v_{\text{rel}}(t) = v_{\text{car}}(t)-v_{\text{truck}}(t) = 20 - (20-4t) = 4t

This starts at $0$ (both at $20\ \text{m/s}$ the instant the truck signals) and rises linearly to $v_{\text{rel}}(0.5) = 2\ \text{m/s}$ — the gap is closing (the car is catching up), and the closing rate is increasing, because the truck is slowing down while the car isn't yet.

Write the closing rate after the car starts braking, $t>0.5\ \text{s}$ :

v_{\text{car}}(t) = 20 - \frac{20}{3}(t-0.5), \qquad v_{\text{rel}}(t) = 4t - \frac{20}{3}(t-0.5)

This is still linear in $t$ (both terms are linear), continuing smoothly from $v_{\text{rel}}(0.5)=2\ \text{m/s}$ , but now decreasing (the car's stronger braking is winning back the closing rate) — until it reaches zero.

Find when the closing stops — i.e. when $v_{\text{rel}}(t)=0$ , the moment the two vehicles' speeds become momentarily equal (this is exactly the instant of closest approach, since the gap can only shrink while $v_{\text{rel}}>0$ and starts widening again once $v_{\text{rel}}<0$ ):

4t = \frac{20}{3}(t-0.5) \;\Longrightarrow\; 12t = 20t-10 \;\Longrightarrow\; t = 1.25\ \text{s}

Recognise the closing-rate graph's shape: two straight segments, forming a simple triangle above the axis. From $t=0$ to $t=0.5$ : rises linearly from $0$ to $2\ \text{m/s}$ . From $t=0.5$ to $t=1.25$ : falls linearly from $2\ \text{m/s}$ back down to exactly $0$ (confirmed by Step 3). The entire closing-rate graph, from $t=0$ to $t=1.25\ \text{s}$ , is a single triangle sitting on the time axis, peaking at $2\ \text{m/s}$ at $t=0.5\ \text{s}$ . …

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