A stationary car is accelerated at the rate of 5 m/s2 . The distance covered by the car till it reaches a speed
of 40 m/s is:
(1) 100 m
(2) 120 m
(3) 160 m
(4) 200 m

---

📘 Problem Statement

A stationary car is accelerated at the rate of 5 m/s².
The distance covered by the car until it reaches a speed of 40 m/s is:

Options:

1. 100 m

2. 120 m

3. 160 m

4. 200 m

---

📚 Concept: Kinematic Equation

To solve this, we use the kinematic equation:

v2=u2+2asv^2 = u^2 + 2asv2=u2+2as

Where:

• $v$ = final velocity = 40 m/s

• $u$ = initial velocity = 0 m/s (since the car is stationary)

• $a$ = acceleration = 5 m/s²

• $s$ = distance covered = ?

---

🧠 Step-by-Step Calculation

Plug the values into the formula:

v2=u2+2as⇒402=0+2×5×s⇒1600=10s⇒s=160010=160 metersv^2 = u^2 + 2as \Rightarrow 40^2 = 0 + 2 \times 5 \times s \Rightarrow 1600 = 10s \Rightarrow s = \frac{1600}{10} = 160 \text{ meters}v2=u2+2as⇒402=0+2×5×s⇒1600=10s⇒s=101600=160 meters

---

✅ Correct Answer: (3) 160 meters

---

📊 Summary Table

---

💡 Why This Problem Matters

This is a classic kinematics question from physics, commonly seen in:

• High school physics exams

• Engineering entrance tests like JEE, NEET, etc.

• Competitive exams with physics sections

It reinforces your understanding of:

• Equations of motion

• Acceleration and velocity relationships

• Solving problems using algebra

---

✅ Key Takeaways

• Use the formula v2=u2+2asv^2 = u^2 + 2asv2=u2+2as when initial speed, final speed, and acceleration are known.

• For a stationary start, set $u=0$.

• Substitute known values and solve for the unknown (distance).