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B. Kinematics · Kinematics: Motion in One and Two Dimensions

Solving Problems with Uniformly Accelerated Motion

B2.3 B2.7 A1.13About 75 minutes
Learning goals
  • I can follow a clear problem-solving plan (Given, Required, Analysis, Solution, Statement).
  • I can choose the right equation of motion by finding the missing variable.
  • I can use signs correctly for vectors in one dimension.
  • I can solve two-stage problems, such as reaction distance plus braking distance.
  • I can solve problems that need a quadratic equation.

Words to know

GRASS
A problem-solving plan: Given, Required, Analysis (choose the equation), Solution, Statement (answer in words with units and direction).
reaction time
The time between seeing a hazard and starting to act (for example, pressing the brake). The vehicle keeps moving at constant velocity during it.
reaction distance
The distance travelled during the reaction time: Δd = vΔt.
braking distance
The distance travelled while the brakes slow the vehicle to a stop.
stopping distance
Reaction distance + braking distance.

From equations to answers

You now have five equations. In real problems the hard part is not the arithmetic. It is deciding which equation to use and keeping the signs straight. A good plan makes that almost automatic.

The GRASS plan

  1. Given: list every known quantity with its symbol, sign and unit. Choose and state the positive direction. Watch for hidden values: "from rest" means v₁ = 0; "comes to a stop" means v₂ = 0.
  2. Required: write the symbol of what you need to find.
  3. Analysis: decide which of Δd, v₁, v₂, a and Δt is not given and not required. Pick the equation that leaves out that variable.
  4. Solution: rearrange first, then substitute numbers with units, then calculate. Keep extra digits until the end.
  5. Statement: answer in a sentence with the correct units, direction and significant digits.
Which variable is NOT given and NOT asked for?no Δd(1)v₂ = v₁ + aΔtno a(2)Δd = ½(v₁ + v₂)Δtno v₂(3)Δd = v₁Δt + ½aΔt²no Δt(4)v₂² = v₁² + 2aΔdno v₁(5)Δd = v₂Δt − ½aΔt²
Choosing an equation. Each equation of uniformly accelerated motion leaves out one of the five variables. Find the one your problem does not involve, and the box below it is your equation.
Worked example

Example 1: How long must the runway be?

A passenger jet starts from rest and accelerates down the runway at 3.5 m/s². It needs a speed of 70 m/s to take off. (a) What minimum runway length is needed? (b) How long does the take-off run last?

  1. Given: forward positive; v₁ = 0, v₂ = +70 m/s, a = +3.5 m/s². Required: (a) Δd, (b) Δt.
  2. Analysis (a): Δt is not given and not asked for in part (a), so use v₂² = v₁² + 2aΔd.
  3. Solution (a): Δd = (v₂² − v₁²) ÷ (2a) = (70² − 0) ÷ (2 × 3.5) = 4900 ÷ 7.0 = 700 m.
  4. Solution (b): now Δd is not needed, so use v₂ = v₁ + aΔt: Δt = (v₂ − v₁) ÷ a = 70 ÷ 3.5 = 20 s.
  5. Statement: the runway must be at least 7.0 × 10² m long, and the take-off run lasts 20 s.

(a) 7.0 × 10² m; (b) 20 s

Worked example

Example 2: Braking to a stop

A car is travelling at 25.0 m/s [N] when the driver brakes. The car stops in 50.0 m. Find (a) its acceleration and (b) the braking time.

  1. Given: north positive; v₁ = +25.0 m/s, v₂ = 0 (stops), Δd = +50.0 m. Required: a, then Δt.
  2. (a) Δt is not given and not asked for: use v₂² = v₁² + 2aΔd. a = (v₂² − v₁²) ÷ (2Δd) = (0 − 625) ÷ (2 × 50.0) = −6.25 m/s².
  3. The minus sign means south (opposite the motion, so the car slows). a = 6.25 m/s² [S].
  4. (b) a is now known, but Equation 2 avoids using a rounded value: Δt = 2Δd ÷ (v₁ + v₂) = 2(50.0) ÷ (25.0 + 0) = 4.00 s.

(a) 6.25 m/s² [S]; (b) 4.00 s

Problems with stages

If the acceleration changes partway through, split the motion into stages. Within each stage the acceleration is constant. The final velocity (and position) of one stage is the starting velocity (and position) of the next.

A very important example is stopping a car. When a driver sees a hazard, there is a reaction time (often taken as about 0.75 s to 1.5 s for an alert driver) before the foot reaches the brake. During that time the car keeps going at full speed. Then the brakes slow it down. Snow and ice greatly reduce how quickly a car can slow down, which is why stopping distances are so much longer in a Canadian winter.

Worked example

Example 3: Stopping distance on a winter road

A driver travelling at 20 m/s (72 km/h) sees a deer on the road. Her reaction time is 0.75 s. Suppose the car can brake at 5.0 m/s² on dry pavement but only 2.5 m/s² on packed snow. Find the total stopping distance in each case.

  1. Stage 1 (reaction): constant velocity, so Δd₁ = vΔt = (20)(0.75) = 15 m. This is the same on any road surface.
  2. Stage 2 (braking, dry): forward positive; v₁ = 20, v₂ = 0, a = −5.0. Use v₂² = v₁² + 2aΔd: Δd₂ = (0 − 400) ÷ (2 × −5.0) = 40 m.
  3. Dry total: 15 + 40 = 55 m.
  4. Stage 2 (braking, snow): a = −2.5: Δd₂ = (0 − 400) ÷ (2 × −2.5) = 80 m.
  5. Snow total: 15 + 80 = 95 m. Halving the acceleration doubles the braking distance.

About 55 m on dry pavement and 95 m on packed snow.

Stopping distance at 20 m/s (72 km/h), drawn to scaledry pavement15 mbraking 40 m= 55 mpacked snow15 mbraking 80 m= 95 m20 mreaction distance (driver has not braked yet)braking distance
Example 3 to scale. The reaction distance is the same on both surfaces, but on snow the braking distance is twice as long. Braking distance also grows with the square of the speed, so going a little faster makes a big difference.
Worked example

Example 4: Two answers from a quadratic

A ball is rolled up a long ramp at 4.0 m/s. Its acceleration is 2.0 m/s² down the ramp. When is the ball 3.0 m up the ramp from its starting point?

  1. Given: up the ramp positive; v₁ = +4.0 m/s, a = −2.0 m/s², Δd = +3.0 m. Required: Δt. v₂ is not given and not asked for: use Δd = v₁Δt + ½aΔt².
  2. 3.0 = 4.0Δt + ½(−2.0)Δt², so 3.0 = 4.0Δt − 1.0Δt².
  3. Rearrange to standard form: Δt² − 4.0Δt + 3.0 = 0. Factor: (Δt − 1.0)(Δt − 3.0) = 0.
  4. Δt = 1.0 s or 3.0 s. Both make sense: the ball passes the 3.0 m mark on the way up at 1.0 s, turns around (at t = 2.0 s, 4.0 m up), and passes the same mark on the way down at 3.0 s.
  5. If a quadratic does not factor, use the quadratic formula from Grade 10. Reject negative times.

At 1.0 s (going up) and at 3.0 s (coming back down).

Avoiding the common traps

  • State the positive direction and give every vector a sign. An object slowing down has an acceleration opposite to its velocity.
  • Units: convert km/h to m/s and minutes to seconds before using the equations.
  • Square roots: Equation 4 gives v₂², so v₂ = ±√(...). Choose the sign from the direction of motion.
  • Rounding: keep all digits until the end, then match the fewest significant digits in the data (A1.13).
  • Check: does the answer make sense? A car does not stop in 0.2 m or take 3 hours to brake.
Just for fun

Some ski jumps put athletes in the air at over 90 km/h. The skis, however, have no brakes. That's why every ski jump ends in a long out-run: a stretch of snow where the jumper has room to slow down, which is a braking-distance problem on skis.

Hands-on activity

Measure your own reaction time, then your 'reaction distance'

You need: a 30 cm ruler and a partner.

  1. Your partner holds the ruler vertically, with the 0 cm mark just between your open thumb and finger. You do not touch the ruler.
  2. Without warning, your partner lets go. Catch it as fast as you can. Read the mark at the top of your thumb.
  3. Repeat 5 times and average. The ruler falls from rest with a = 9.8 m/s², so your reaction time is Δt = √(2Δd ÷ 9.8), with Δd in metres.
  4. Calculate how far a car at 20 m/s (72 km/h) would travel during your reaction time.
  5. Repeat while a family member talks to you or asks you a question. What happens?

What to look for: a catch at 18 cm gives Δt = √(2 × 0.18 ÷ 9.8) ≈ 0.19 s; at 20 m/s that is about 3.8 m. Real driving reactions are longer than this lab test (you must notice the hazard, decide, and move your foot), which is why 0.75 s or more is used. Distraction usually makes the catch distance noticeably larger. (This is why texting while driving is so dangerous.)

Your turn

1

A cyclist starts from rest and accelerates at 1.2 m/s² for 5.0 s. How far does she travel?

2

A car accelerates from 15 m/s to 27 m/s over 84 m. Find its acceleration.

3

A train slowing at 0.50 m/s² comes to a stop in 40 s. How fast was it going?

4

For the same train, how far did it travel while stopping?

5

A puck slides 18 m on ice and slows from 8.0 m/s to 4.0 m/s. Find the time.

6

A car moving at 90 km/h brakes with an acceleration of 6.0 m/s² opposite its motion. Find the braking distance.

7

A sprinter reaches 10 m/s after 3.0 s of constant acceleration from rest. How far did she run in that time?

More practice: warm-up

1

What is v₁ for an object that starts "from rest"?

2

What is v₂ for an object that "comes to a stop"?

3

Which equation leaves out Δd?

4

From rest, a = 2.0 m/s². Speed after 6.0 s?

5

Convert 54 km/h to m/s.

More practice: on your own

1

A snowmobile at 12 m/s [E] accelerates at 1.5 m/s² [E] for 120 m. Find its final velocity.

2

A ball rolls up a ramp at 3.0 m/s and slows at 1.5 m/s². How far up does it get before stopping?

3

A driver going 30 m/s has a reaction time of 1.0 s and brakes at 6.0 m/s². Find the stopping distance.

4

A car travels 125 m while slowing uniformly to 5.0 m/s over 10 s. Find its starting speed and its acceleration.

5

A cart at +2.0 m/s has a = −1.0 m/s². When is it 1.5 m from the start in the positive direction?

Challenge

1

A speeder passes a parked police car at a constant 30 m/s. At that instant the police car starts from rest with a = 4.0 m/s². When and where does the police car catch up, and how fast is it going then?

2

By what factor does braking distance change if speed doubles (same acceleration)? Use an equation to explain.

3

A ball rolled up a ramp at 5.0 m/s has a = −2.0 m/s². Use the quadratic formula to find when it is 4.0 m up the ramp. Give both times.

Check your understanding

Quiz

1.

You know v₁, v₂ and Δd. Which equation gives a directly?

2.

From rest, a car accelerates at 2.5 m/s² for 8.0 s. Its final speed is:

3.

A car at 20 m/s brakes at 4.0 m/s². The braking distance is:

4.

During a driver's reaction time, the car:

5.

If a car's speed doubles, its braking distance (same braking acceleration):

6.

Solving a motion problem gives Δt = 2.0 s or Δt = −0.5 s. The best response is:

For parents and teachers

Big idea: This lesson turns the equations (B2.3) into a reliable problem-solving method for one-dimensional motion (B2.7), with careful significant digits (A1.13). The stopping-distance example is a real safety lesson for new drivers.

How to help: Ask your teen to say the plan out loud: "What's given? What's required? Which variable is missing?" Look over solutions for a stated positive direction and units on every line.

Common difficulties:

  • Picking an equation before listing the variables.
  • Missing hidden information ("from rest", "stops").
  • Sign errors: an object slowing down needs an acceleration with the opposite sign to its velocity.
  • Forgetting the reaction stage in stopping problems.
  • Rejecting a valid second answer in quadratic problems (the object may pass the same point twice).