9478 / 2027
Motion and Forces overview

Lesson 3 of 6

Derive and use constant-acceleration equations

  • I can derive straight-line constant-acceleration equations from the definitions of velocity and acceleration.Syllabus

    Syllabus 9478, 3(d). Derive straight-line constant-acceleration equations from the definitions of velocity and acceleration.

  • I can solve straight-line constant-acceleration problems, including vertical motion without air resistance in a uniform gravitational field.Syllabus

    Syllabus 9478, 3(e). Solve straight-line constant-acceleration problems, including vertical motion without air resistance in a uniform gravitational field.

The familiar motion equations follow from constant acceleration along one straight line. Check that model before substituting numbers.

Start the chosen interval at t = 0. Let u be its initial velocity, v its final velocity, a the constant acceleration, t its duration and s its signed displacement. Use the same positive direction for every signed value. A reversal on the same line is allowed.

Build the equations from a rate and a graph area

  1. Start with acceleration.

    For constant acceleration, a = (v - u)/t. Rearranging gives v = u + at. Velocity therefore changes linearly with time.

  2. Use the straight velocity-time line.

    Its signed area gives displacement: s = (u + v)t/2. The mean of the endpoint velocities is the average velocity because the graph is straight. If it crosses zero, separate positive and negative areas give the same signed result.

  3. Eliminate final velocity.

    Substitute v = u + at into the area equation:
    s = [u + (u + at)]t/2
    = ut + ½at2.

  4. Alternatively, eliminate initial velocity.

    Use u = v - at:
    s = [(v - at) + v]t/2
    = vt - ½at2.

  5. Eliminate time.

    Multiply v - u = at by v + u:
    v2 - u2 = a(v + u)t.
    Since (v + u)t = 2s, this gives v2 = u2 + 2as. This derivation does not require division by a and remains consistent when a = 0.

Choose an equation that connects your known quantities to the required one. For stages with different constant accelerations, start a new interval for each stage and carry its ending position and velocity into the next. Do not use one acceleration across a stage where it changes.

Check your understanding

A body moves along a straight line with a curved velocity-time graph. Why is displacement = (initial velocity + final velocity) x time / 2 generally unsuitable?

  1. Using the average acceleration in a constant-acceleration formula fixes the problem for this motion.
  2. The formula cannot be used for any motion in which the velocity changes.
  3. The endpoint mean equals average velocity just for a straight velocity-time graph.
Show the answer

The endpoint mean equals average velocity just for a straight velocity-time graph.

The derivation uses constant acceleration and a straight velocity-time line. For the curved graph, determine the actual signed area or an appropriate numerical estimate.

A reversal is still straight-line motion

Displacement need not have the final velocity's sign

Take u = +6.0 m/s, a = -2.0 m/s2 and t = 5.0 s.

v = 6 + (-2)(5) = -4 m/s.
s = 6(5) + ½(-2)(52) = +5 m.

For distance, locate the turn first: 0 = 6 - 2t gives t = 3 s. The body travels 9 m before the turn and 4 m afterwards, for 13 m total. The +5 m result from the equation is displacement.

Keep one sign convention during vertical motion

An object is launched vertically upward at 12.0 m/s from a point 1.50 m above a floor. Take upward as positive and the release point as y = 0. Neglect air resistance and use uniform g = 9.81 m/s2, so a = -9.81 m/s2 throughout the flight.

Keep the release origin and upward-positive convention

Launch upwards from one and a half metres above the floorHeights use forty drawing units per metre, with release at y equals zero and the floor at minus 1.50 metres. The point-like object is launched at positive twelve metres per second. Its highest point is 7.33945 metres above release, at about 1.22 seconds, with instantaneous velocity zero. Acceleration is negative 9.81 metres per second squared throughout. The floor is reached after about 2.57 seconds with downward velocity negative 13.2 metres per second. Velocity and acceleration arrows are drawn beside the one-dimensional path, with schematic lengths; they do not compare unlike quantities. The downward velocity arrow ends beside the floor without depicting any motion after impact.Top: v = 0; t = 1.22 s7.34 mauv1.50 mRelease: y = 0Floor: y = -1.50 mJust before impact: v = -13.2 m/s+ up

The height scale is shared; the coloured arrows show directions schematically. Here u = +12.0 m/s and a = -9.81 m/s2, including at the top. The impact calculation uses displacement -1.50 m, not the total path travelled.

The release point is the position origin, so the floor is at -1.50 m. Velocity is zero at the top, but acceleration remains downward on ascent, at the top and on descent.

Worked vertical motion

Find the height, then choose the impact roots

  1. Time to the top: v = 0, so 0 = 12.0 - 9.81t. Thus t = 1.223... s, or 1.22 s.
  2. Rise above release: 0 = 12.02 + 2(-9.81)s. Thus s = 7.339... m, or 7.34 m. Adding the initial height gives 8.84 m above the floor.
  3. Velocity just before floor impact: s = -1.50 m, so v2 = 12.02 + 2(-9.81)(-1.50) = 173.43 m2/s2. The object is moving downward, so select v = -13.2 m/s.
  4. Time to impact: -1.50 = 12.0t - 4.905t2, or 4.905t2 - 12.0t - 1.50 = 0. The quadratic formula gives t = [12.0 ± √(12.02 + 4(4.905)(1.50))]/9.81. The roots are 2.5657... s and -0.1192... s. The required post-launch event is t = 2.57 s; the negative root is outside the stated t ≥ 0 interval.

Try it step by step

Work it out in steps: Time to the top (v = 0) t = u / g in s, then Maximum height s = u2 / (2g) in m, then Velocity on return to the launch point in m/s.

  1. Problem 1. A ball is thrown vertically upwards at 15 m/s. Take g = 9.81 m/s2 and upwards as positive; ignore air resistance. Find the time to the top, the maximum height and the velocity when it returns to the launch point.
  2. Problem 2. A ball is thrown vertically upwards at 20 m/s. Take g = 9.81 m/s2 and upwards as positive; ignore air resistance. Find the time to the top, the maximum height and the velocity on return.
  3. Problem 3. A ball is thrown vertically upwards at 8.0 m/s. Take g = 9.81 m/s2 and upwards as positive; ignore air resistance. Find the time to the top, the maximum height and the velocity on return.
Show the answer

Problem 1.

Time to the top (v = 0) t = u / g in s: t = 15 / 9.81 = 1.53 s

Maximum height s = u2 / (2g) in m: s = 152 / (2 x 9.81) = 11.5 m

Velocity on return to the launch point in m/s: v = -15 m/s: same speed, downwards (v2 = u2 + 2as with s = 0)

Problem 2.

Time to the top (v = 0) t = u / g in s: t = 20 / 9.81 = 2.04 s

Maximum height s = u2 / (2g) in m: s = 202 / (2 x 9.81) = 20.4 m

Velocity on return to the launch point in m/s: v = -20 m/s (downwards)

Problem 3.

Time to the top (v = 0) t = u / g in s: t = 8.0 / 9.81 = 0.815 s

Maximum height s = u2 / (2g) in m: s = 8.02 / (2 x 9.81) = 3.26 m

Velocity on return to the launch point in m/s: v = -8.0 m/s (downwards)

Squaring velocity removes its sign; the physical direction selects the root. Zero velocity at the highest point does not mean zero acceleration or zero resultant force.

The model ends just before floor contact. During impact the contact force changes the acceleration. Significant air resistance would also undermine the gravity-only constant-acceleration model.

Check your understanding

An object is launched upward from above a floor. Upward is positive. At its later floor impact, an equation gives v2 = 173.43 m2/s2. Which velocity is appropriate?

  1. -13.2 m/s, because the impact motion is downward.
  2. 0 m/s, because the floor stops the object.
  3. +13.2 m/s, because a square root is positive.
Show the answer

-13.2 m/s, because the impact motion is downward.

sqrt(173.43) is about 13.17 m/s. The physical direction selects the negative root, giving -13.2 m/s to three significant figures.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A car travels at 25 m s-1. The driver sees a hazard and brakes after a reaction time of 0.70 s. The brakes give a constant deceleration of 6.5 m s-2. Calculate the total stopping distance. [4]

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