Chapter 7: Motion Application Questions | science Class 9 CBSE

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Chapter 7: Motion Application Questions | science Class 9 CBSE | ByteNotes.in

Application Question

Concept in a practical situation

Application

Question 1

Applied Concept

Q1. Define average speed and average velocity with examples.

Average speed = total distance / total time.
Average velocity = total displacement / total time.
If object returns to start, avg. velocity = 0, but speed ≠ 0.
E.g., round-trip driving: 100 km to and fro in 2 hr → speed = 100 km/h, velocity = 0.
Speed is scalar; velocity is vector.

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Application Question

Concept in a practical situation

Application

Question 2

Applied Concept

Q2. Derive the equation: v = u + at.

Let initial velocity = u, acceleration = a, time = t.
Acceleration = (v - u)/t → v = u + at.
v = final velocity after time t.
This equation connects initial/final velocity with time.
Applies to uniform acceleration only.

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Application Question

Concept in a practical situation

Application

Question 3

Applied Concept

Q3. Derive the second equation: s = ut + ½ at².

Let initial velocity = u, acceleration = a, time = t.
Final velocity = v = u + at.
Average velocity = (u + v)/2 = (u + u + at)/2 = u + at/2.
Distance = avg. velocity × time = (u + at/2) × t = ut + ½ at².
Applies to uniformly accelerated motion.

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Application Question

Concept in a practical situation

Application

Question 4

Applied Concept

Q4. Derive the third equation: v² = u² + 2as.

Use: v = u + at and s = ut + ½ at².
Eliminate time: t = (v - u)/a.
Substitute in s: s = [(u + v)/2] × t → 2as = v² - u².
Rearranging gives v² = u² + 2as.
Useful when time is not given.

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Application Question

Concept in a practical situation

Application

Question 5

Applied Concept

Q5. A car accelerates from 5 m/s to 25 m/s in 4 s. Find acceleration and distance covered.

u = 5 m/s, v = 25 m/s, t = 4 s.
a = (v - u)/t = (25 - 5)/4 = 5 m/s².
s = ut + ½ at² = 5×4 + ½×5×16 = 20 + 40 = 60 m.
Acceleration = 5 m/s².
Distance = 60 m.

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Application Question

Concept in a practical situation

Application

Question 6

Applied Concept

Q6. An object moves at 10 m/s for 10 s, then stops in next 5 s. Find acceleration during deceleration.

Initial speed = 10 m/s, final = 0, time = 5 s.
a = (0 - 10)/5 = -2 m/s².
Negative sign shows retardation.
Uniform deceleration assumed.
Retardation = 2 m/s².

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Chapter sections

Chapter 7: Motion | Application Questions

Q1. Define average speed and average velocity with examples.

Q2. Derive the equation: v = u + at.

Q3. Derive the second equation: s = ut + ½ at².

Q4. Derive the third equation: v² = u² + 2as.

Q5. A car accelerates from 5 m/s to 25 m/s in 4 s. Find acceleration and distance covered.

Q6. An object moves at 10 m/s for 10 s, then stops in next 5 s. Find acceleration during deceleration.

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