Force from Speed & Distance — NTSE Physics in 59 Sec ⚡

 

❓ Question

A vehicle starting from rest attains a speed of 72 km/h after covering a distance of 100 m.
If the mass of the vehicle is 500 kg, find the force exerted by the vehicle.


🖼️ Concept Image

Force from Speed & Distance — NTSE Physics in 59 Sec ⚡


✍️ Short Solution

This is a pure kinematics + Newton’s second law question.
No tricks, no time calculation — just v² = u² + 2as and F = ma 💯


🔹 Step 1 — Convert Speed into SI Units (MOST IMPORTANT 💯)

Given speed is in km/h, but formulas need m/s.

72 km/h=20 m/s

📌 NTSE aur school exams mein yahin sabse zyada silly mistakes hoti hain.


🔹 Step 2 — Write Known Values Clearly

Initial speed:

u=0(starting from rest)u = 0 \quad \text{(starting from rest)}

Final speed:

v=20 m/sv = 20 \text{ m/s}

Distance covered:

s=100 ms = 100 \text{ m}

Mass of vehicle:

m=500 kgm = 500 \text{ kg}

🔹 Step 3 — Use Equation of Motion

Since u, v, s are given, use:

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

Substitute values:

(20)2=0+2a(100)(20)^2 = 0 + 2a(100)
400=200a400 = 200a
a=2 m/s2a = 2 \text{ m/s}^2

📌 Acceleration mil gaya — question ab almost done hai ✅


🔹 Step 4 — Apply Newton’s Second Law

Force is given by:

F=maF = ma
F=500×2F = 500 \times 2
F=1000 NF = 1000 \text{ N}

✅ Final Answer

F=1000 N\boxed{F = 1000\ \text{N}}



⭐ Golden NTSE Insight

  • Jab speed + distance given ho →
    v² = u² + 2as sabse powerful formula hai

  • Time nikaalne ki koi zarurat nahi

  • Force ke liye hamesha:

F=maF = ma

🧠 One-line memory:

Speed aur distance mile — time ko ignore karo.

Comments

Popular posts from this blog

Ideal Gas Equation Explained: PV = nRT, Units, Forms, and JEE Tips [2025 Guide]

Balanced Redox Reaction: Mg + HNO₃ → Mg(NO₃)₂ + N₂O + H₂O | JEE Chemistry

If the equation of the line passing through the point (0, −1/2, 0) and perpendicular to the lines r = λ(î + aĵ + bk̂) and r = ( î - ĵ - 6k̂) + μ(−bî + aĵ + 5k̂) is x-1/-2 = y+4/d = z-c/-4, then a + b + c + d is equal to :