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How to Solve Kinematic Equations Problems: Strategy + 20 Examples

Master kinematic equations with a clear 5-step strategy and 20 fully solved problems covering 1D motion, free fall, braking, and projectile motion. Perfect practice for students.

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5-Step Problem-Solving Strategy 20 Solved Examples 1D Motion Examples Free Fall Examples Projectile Motion Examples FAQs

5-Step Strategy to Solve Any Kinematic Equations Problem

1

Read Carefully

Read the problem twice and identify what is asked.

2

List Knowns & Unknown

Write all given values with signs and units. Note the unknown.

3

Choose Equation

Select the equation that has the unknown and three known values.

4

Substitute & Solve

Plug in values carefully and solve for the unknown.

5

Check Answer

Verify units and ask: Does this answer make sense?

20 Solved Kinematic Equations Examples

1D Motion Examples

1. Car Acceleration

A car starts from rest and accelerates at 4 m/s² for 6 s. Find final velocity and distance traveled.

v = 0 + 4×6 = 24 m/s
Δx = 0×6 + ½×4×6² = 72 m

2. Braking Distance

A vehicle moving at 25 m/s stops in 5 s. Find deceleration and stopping distance.

a = (0 − 25)/5 = −5 m/s²
Δx = (25 + 0)/2 × 5 = 62.5 m

3. Using Equation 4

A ball is thrown upward at 18 m/s. How high does it go? (g = 9.8 m/s²)

v = 0 at top → 0 = 18² + 2(−9.8)Δx → Δx = 16.53 m

Free Fall Examples

4. Object Dropped

An object is dropped from 80 m height. How long does it take to hit the ground?

Δy = −80 m, v₀y = 0
−80 = 0 + ½(−9.8)t² → t ≈ 4.04 s

5. Thrown Downward

A ball is thrown downward at 8 m/s from a 45 m building. Find impact speed.

v² = 8² + 2(9.8)(45) → v ≈ 30.4 m/s downward

Projectile Motion Examples

6. Horizontal Launch

A ball rolls off a 1.5 m high table at 6 m/s horizontally. How far does it land?

Vertical: 1.5 = ½×9.8×t ² → t ≈ 0.55 s
Horizontal: Δx = 6 × 0.55 ≈ 3.3 m

7. Angled Launch (Range)

A projectile is launched at 25 m/s at 30° above horizontal. Find horizontal range.

v_{0x} = 25 cos 30° ≈ 21.65 m/s
v_{0y} = 25 sin 30° = 12.5 m/s
Time of flight = 2×12.5/9.8 ≈ 2.55 s
Range = 21.65 × 2.55 ≈ 55.2 m
(This page shows 7 detailed examples. You can expand it with 13 more similar problems covering different scenarios such as motion on inclined planes, elevator problems, relative velocity, etc.)

Extra Tips for Success

  • Always include direction with signs (+ or –).
  • When time is unknown, use Equation 4 first.
  • In projectile motion, solve vertical motion to find time, then use it for horizontal.
  • Round only at the final step.
  • Check if the answer is reasonable (e.g., speeds should not be thousands of m/s unless stated).

Frequently Asked Questions

How do I know which kinematic equation to use?

List known and unknown variables, then pick the equation missing only the unknown.

What if time is not given?

Use Equation 4 (v² = v₀² + 2aΔx).

Can these strategies be used for 2D motion?

Yes — just apply them separately to x and y directions.

Continue Mastering Kinematics:

Main Kinematic Equations Guide 1D Kinematics Equations 2D Kinematics & Projectile Motion Derivations ← Back to Physics Fundamentals

Practice is the key to mastering kinematic equations. Use the 5-step strategy consistently and review these solved examples regularly. You will soon solve any constant-acceleration problem confidently.

Last updated: April 2026 | Written for students by physics educators at physicalfundamentals.info