Fisica

Uniformly Accelerated Motion: Solved Exercises and Formulas

The formulas of uniformly accelerated motion explained and applied: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s, with five solved and verified exercises.

Recommended for: Grade 9 · Grade 10

In uniformly accelerated motion an object moves along a straight line with constant acceleration: its velocity changes by the same amount every second. The three relations you need are v = v₀ + a · t, s = s₀ + v₀ · t + ½ · a · t² and v² = v₀² + 2 · a · s, where v₀ is the initial velocity and a is the acceleration.

The three formulas and when to use them

The first one, v = v₀ + a · t, tells you how the speed grows: a car accelerating at 3 m/s² gains 3 m/s every second. The second, s = v₀ · t + ½ · a · t², gives the distance covered and simplifies to s = ½ · a · t² when the object starts from rest. The third, v² = v₀² + 2 · a · s, is the one that saves you when time never appears in the problem.

Choosing between them is almost automatic: look at which quantity is missing from the question. If the distance is missing, use the first; if the final speed is missing, use the second; if the time is missing, use the third.

Braking and negative acceleration

Braking is uniformly accelerated motion with the acceleration pointing opposite to the velocity. In practice you pick a positive direction (usually the direction of travel), write the velocity with a plus sign and the acceleration with a minus sign, then let the algebra do the work. If your time or distance comes out negative, you almost certainly flipped a sign at the start.

Watch out for one detail that trips people up: braking distance grows with the square of the speed. Doubling the speed quadruples the distance needed to stop, and you can read that straight off v² = v₀² + 2 · a · s.

Reading the graphs

On a velocity–time graph, uniformly accelerated motion is a sloping straight line, and its slope is the acceleration. On a position–time graph it is a parabola that gets steeper and steeper, because the object covers more ground with every passing second.

A handy trick for tests: the area under the velocity–time graph is the distance travelled. If that area is a trapezium you can work out the distance without touching the equation of motion, which gives you an independent way to check your answer.

A method that avoids mistakes

Start by listing the data with their units (metres, seconds, m/s, m/s²) and convert any km/h by dividing by 3.6. Then mark the unknown and pick the formula that does not contain the missing quantity. Finally, substitute the numbers only at the end, once the unknown is isolated.

The exercises below apply this method at increasing difficulty, from starting from rest to one object chasing another. Every result has been re-checked with a second formula.

Solved exercises

1. A car starts from rest and accelerates uniformly at 3 m/s². What speed does it reach after 8 s? base

Show solution
  1. The car starts from rest, so v₀ = 0.
  2. Use the velocity equation: v = v₀ + a · t = 0 + 3 · 8.
  3. v = 24.

Answer: v = 24 m/s

2. A scooter starts from rest with a constant acceleration of 2 m/s². How far does it travel in 6 s? base

Show solution
  1. With v₀ = 0 the equation of motion reduces to s = ½ · a · t².
  2. Substitute: s = ½ · 2 · 6² = ½ · 2 · 36.
  3. s = 36.

Answer: s = 36 m

3. A train is moving at 20 m/s and accelerates uniformly at 1.5 m/s² for 10 s. Find its final speed and the distance covered. intermedio

Show solution
  1. Final speed: v = v₀ + a · t = 20 + 1.5 · 10 = 35 m/s.
  2. Distance: s = v₀ · t + ½ · a · t² = 20 · 10 + ½ · 1.5 · 100.
  3. s = 200 + 75 = 275 m.
  4. Check with v² = v₀² + 2·a·s: 35² − 20² = 1225 − 400 = 825 and 2 · 1.5 · 275 = 825. It matches.

Answer: v = 35 m/s ; s = 275 m

4. A car travelling at 30 m/s brakes uniformly and stops after 75 m. What is its acceleration, and how long does the braking last? intermedio

Show solution
  1. The time is unknown, so use the time-free formula: v² = v₀² + 2 · a · s.
  2. At the stop v = 0: 0 = 30² + 2 · a · 75, that is 0 = 900 + 150 · a.
  3. a = −900 / 150 = −6 m/s²: the minus sign means it is decelerating.
  4. Time: from v = v₀ + a · t, t = (0 − 30) / (−6) = 5 s.
  5. Check: s = 30 · 5 + ½ · (−6) · 5² = 150 − 75 = 75 m. It matches.

Answer: a = −6 m/s² ; t = 5 s

5. A car waiting at a traffic light pulls away with a constant acceleration of 2 m/s² at the exact moment a truck passes it at a constant 12 m/s. How long does it take the car to catch the truck, and how far from the traffic light does this happen? avanzato

Show solution
  1. Write both equations of motion with the origin at the traffic light and the same starting instant.
  2. Car (accelerating from rest): s₁ = ½ · 2 · t² = t².
  3. Truck (uniform motion): s₂ = 12 · t.
  4. They meet when s₁ = s₂: t² = 12 · t, that is t · (t − 12) = 0.
  5. The solution t = 0 is the starting instant, so they actually meet at t = 12 s.
  6. Distance: s = 12² = 144 m (and indeed 12 · 12 = 144 m for the truck too).
  7. Useful check: at that moment the car is doing v = 2 · 12 = 24 m/s, twice the truck's speed.

Answer: t = 12 s ; s = 144 m from the traffic light

FAQ

When should I use the formula v² = v₀² + 2·a·s?

Whenever the problem neither gives you the time nor asks for it: it is the only relation linking speed, acceleration and distance without t appearing. It is the standard shortcut for braking problems, where you know the initial speed and the stopping distance.

Does a negative acceleration mean the object moves backwards?

No. The sign of the acceleration only tells you its direction relative to the axis you chose. If acceleration and velocity have opposite signs the object slows down; if they have the same sign it speeds up. A braking car has positive velocity and negative acceleration, yet it keeps moving forward until the velocity reaches zero.

How do the graphs of uniform and uniformly accelerated motion differ?

In uniform motion the position–time graph is a straight line and the velocity–time graph is horizontal. In uniformly accelerated motion the velocity–time graph is a sloping straight line (its slope is the acceleration) and the position–time graph is a parabola. In both cases, the area under the velocity–time graph is the distance travelled.