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📚 Class 9 Science | Chapter 4 | Exploration NCERT

Activity 4.2 — Average Acceleration of Cars (0 to 100 km h⁻¹)

Car advertisements boast "0 to 100 in X seconds" — but what does that mean in physics? This activity converts those times directly into acceleration in m s⁻², making the concept real and relatable. From NCERT Chapter 4 (Exploration edition) Class 9 Science. Aligned with CBSE syllabus 2026-27.

100 km h⁻¹ = 27.78 m s⁻¹
a = (v − u) / t
Shorter time → higher acceleration

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1. Aim and Pre-requisite

Aim: To calculate the average acceleration of different cars given their time to accelerate from 0 to 100 km h⁻¹, and to compare these accelerations.

Pre-requisite — Convert 100 km h⁻¹ to m s⁻¹:

All acceleration calculations in SI units require velocity in m s⁻¹. The speed 100 km h⁻¹ must be converted once and used throughout the table:

$$v = 100\,\text{km h}^{-1} = 100 \times \frac{5}{18} = \frac{500}{18} \approx 27.78\,\text{m s}^{-1}$$

Since all cars start from rest, the initial velocity $u = 0$ in every case. The only variable that changes from car to car is the time $t$ taken to reach 27.78 m s⁻¹.


2. Formula and Unit

The formula for average acceleration (Eq. 4.3c from NCERT Chapter 4) is:

$$a = \frac{v - u}{t_2 - t_1} = \frac{v - u}{t}$$

In this activity: $u = 0$, $v = 27.78\,\text{m s}^{-1}$, $t$ = time given for each car. Therefore:

$$a = \frac{27.78 - 0}{t} = \frac{27.78}{t}\,\text{m s}^{-2}$$ where $t$ is the 0–100 time of the car in seconds

The SI unit of acceleration is metre per second squared (m s⁻²). A larger value of $a$ means the car reaches 100 km h⁻¹ faster — it is more "sporty." A smaller $a$ means the car takes longer and accelerates more gently.


3. Completed Table 4.2 — Sample Data

NCERT asks students to look up (or use given) 0–100 km h⁻¹ times for different cars and fill Table 4.2. The following typical values are used as representative examples:

Car Type Time t (s) v (m s⁻¹) u (m s⁻¹) Acceleration a (m s⁻²)
Economy hatchback 12 27.78 0 2.31
Mid-size sedan 8 27.78 0 3.47
Performance SUV 6 27.78 0 4.63
Sports car 4 27.78 0 6.95

Sample calculation (for economy hatchback, t = 12 s):

$$a = \frac{27.78 - 0}{12} = \frac{27.78}{12} \approx 2.31\,\text{m s}^{-2}$$

Sample calculation (for sports car, t = 4 s):

$$a = \frac{27.78}{4} \approx 6.95\,\text{m s}^{-2}$$
Note for students: NCERT Activity 4.2 asks you to look up actual car specifications from newspapers, magazines, or websites, and fill in the table with real data. The values above are illustrative. The acceleration you calculate will depend on the car you choose. The formula and method remain the same regardless of which car data you use.

4. Observations and Conclusions

Q. What pattern do you observe in the table?

As the 0–100 time decreases, the acceleration increases. This is a direct consequence of the formula $a = 27.78/t$ — acceleration and time are inversely related when the velocity change is fixed. A sports car achieving the same 100 km h⁻¹ in one-third the time of an economy car has three times the acceleration.

Q. All four cars reach the same final speed. Why do their accelerations differ?

Acceleration is rate of change of velocity — not the final velocity itself. Since all cars change velocity by the same amount (0 to 27.78 m s⁻¹), the one that does it in less time has a higher rate of change, hence higher acceleration. A powerful engine can exert a larger force on the car, causing greater acceleration.

Q. Is the acceleration in this activity constant or average?

It is average acceleration — the formula $a = (v-u)/t$ gives the net change in velocity over the total time, without saying anything about how velocity changed moment to moment. In reality, a car's acceleration varies as it moves through gears and as air resistance increases. The value from the 0–100 test is a useful single-number summary but is not the instantaneous acceleration at any particular moment.

Connection to acceleration due to gravity: Notice that even a sports car at ~7 m s⁻² has acceleration less than $g = 9.8\,\text{m s}^{-2}$. Free-fall gives more acceleration than most road cars — a useful benchmark to remember from Average Acceleration (C03).



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5. Practice Questions

Q1. A car goes from 0 to 100 km h⁻¹ in 10 s. Calculate its average acceleration in m s⁻².

Show Solution
$v = 100 \times 5/18 = 27.78\,\text{m s}^{-1}$; $u = 0$; $t = 10\,\text{s}$

$$a = \frac{27.78 - 0}{10} = \mathbf{2.78\,\text{m s}^{-2}}$$

Q2. Car A does 0–100 in 5 s and Car B in 15 s. How many times greater is Car A's acceleration than Car B's?

Show Answer
$a_A = 27.78/5 = 5.56\,\text{m s}^{-2}$; $a_B = 27.78/15 = 1.85\,\text{m s}^{-2}$

Ratio = $5.56/1.85 = \mathbf{3}$. Car A has three times the acceleration of Car B — exactly the ratio of their times (15/5 = 3), confirming the inverse relationship.

Q3. An electric car accelerates from 0 to 100 km h⁻¹ in 3.5 s. Express this acceleration as a multiple of g (g = 9.8 m s⁻²). What does this tell you?

Show Answer
$a = 27.78/3.5 = 7.94\,\text{m s}^{-2}$. As a multiple of $g$: $7.94/9.8 \approx 0.81g$.

This means the car accelerates at about 81% of free-fall acceleration. The driver experiences a force pushing them back into their seat of about 0.81 times their own weight — quite noticeable and the upper limit of what most road cars can achieve.

📚 More from Chapter 4 — Describing Motion Around Us


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