Constant acceleration and the equations of motion

Key ideas

When acceleration is constant, five quantities describe the whole motion. The exam resource sheet gives you four equations linking them — each one is missing exactly one variable, which is how you choose between them.

Symbol	Quantity	Unit
$v_i$	initial velocity	m s⁻¹
$v_f$	final velocity	m s⁻¹
$a$	acceleration	m s⁻²
$d$	distance	m
$t$	time	s

Equation	Doesn't contain
$v_f = v_i + at$	$d$
$d = v_i t + \tfrac{1}{2}at^2$	$v_f$
$v_f^2 = v_i^2 + 2ad$	$t$
$d = \dfrac{v_i + v_f}{2}\,t$	$a$

Strategy: list what you know and what you want, then pick the equation that skips the variable you neither know nor want.

Worked ExampleChoosing the right equation of motion

A car accelerates from $5.0$ m s⁻¹ at $2.0$ m s⁻² for $4.0$ s. Find its final velocity and the distance travelled.

Step 1 — List what you know and want

Known: $v_i = 5.0$ , $a = 2.0$ , $t = 4.0$ . Wanted: $v_f$ , then $d$ .

Step 2 — Final velocity (the equation with no $d$ )

$v_f = v_i + at = 5.0 + (2.0)(4.0) = 13 \text{ m s}^{-1}$

Step 3 — Distance (the equation with no $v_f$ )

$d = v_i t + \tfrac{1}{2}at^2 = (5.0)(4.0) + \tfrac{1}{2}(2.0)(4.0)^2 = 20 + 16 = 36 \text{ m}$

Free fall

An object in free fall has constant acceleration $g = 9.8$ m s⁻² downward, and the same four equations apply. A ball dropped from rest ( $v_i = 0$ ) off a bridge falls for $1.5$ s:

$v_f = 0 + (9.8)(1.5) = 14.7 \approx 15 \text{ m s}^{-1}$

$d = 0 + \tfrac{1}{2}(9.8)(1.5)^2 = 11 \text{ m (2 s.f.)}$

Practice question

A cyclist travelling at $12$ m s⁻¹ brakes and stops in a distance of $36$ m. Find her acceleration.

Worked solution: know $v_i = 12$ , $v_f = 0$ , $d = 36$ ; want $a$ ; no time involved, so use $v_f^2 = v_i^2 + 2ad$ : $0 = 144 + 2a(36)$ , so $a = -\dfrac{144}{72} = -2.0$ m s⁻². The negative sign means the acceleration is opposite to the motion (deceleration).

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A cyclist starts from rest and accelerates uniformly, reaching a speed of $8.0$ m s⁻¹ after $4.0$ s.

Show that her acceleration is $2.0$ m s⁻².

Merit

A car travelling at $6.0$ m s⁻¹ accelerates at $3.0$ m s⁻² over a distance of $24$ m.

Calculate the car's final velocity.

Excellence

A stone is thrown vertically upward at $15$ m s⁻¹ (take $g = 9.8$ m s⁻² downward).

Calculate the maximum height the stone reaches, and explain the sign convention you used and why the velocity is zero at the highest point.

Constant acceleration and the equations of motion

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Key ideas

Symbol	Quantity	Unit
$v_i$	initial velocity	m s⁻¹
$v_f$	final velocity	m s⁻¹
$a$	acceleration	m s⁻²
$d$	distance	m
$t$	time	s

Equation	Doesn't contain
$v_f = v_i + at$	$d$
$d = v_i t + \tfrac{1}{2}at^2$	$v_f$
$v_f^2 = v_i^2 + 2ad$	$t$
$d = \dfrac{v_i + v_f}{2}\,t$	$a$

Strategy: list what you know and what you want, then pick the equation that skips the variable you neither know nor want.

Worked ExampleChoosing the right equation of motion

A car accelerates from $5.0$ m s⁻¹ at $2.0$ m s⁻² for $4.0$ s. Find its final velocity and the distance travelled.

Step 1 — List what you know and want

Known: $v_i = 5.0$ , $a = 2.0$ , $t = 4.0$ . Wanted: $v_f$ , then $d$ .

Step 2 — Final velocity (the equation with no $d$ )

$v_f = v_i + at = 5.0 + (2.0)(4.0) = 13 \text{ m s}^{-1}$

Step 3 — Distance (the equation with no $v_f$ )

$d = v_i t + \tfrac{1}{2}at^2 = (5.0)(4.0) + \tfrac{1}{2}(2.0)(4.0)^2 = 20 + 16 = 36 \text{ m}$

Free fall

An object in free fall has constant acceleration $g = 9.8$ m s⁻² downward, and the same four equations apply. A ball dropped from rest ( $v_i = 0$ ) off a bridge falls for $1.5$ s:

$v_f = 0 + (9.8)(1.5) = 14.7 \approx 15 \text{ m s}^{-1}$

$d = 0 + \tfrac{1}{2}(9.8)(1.5)^2 = 11 \text{ m (2 s.f.)}$

Practice question

A cyclist travelling at $12$ m s⁻¹ brakes and stops in a distance of $36$ m. Find her acceleration.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A cyclist starts from rest and accelerates uniformly, reaching a speed of $8.0$ m s⁻¹ after $4.0$ s.

Show that her acceleration is $2.0$ m s⁻².

Merit

A car travelling at $6.0$ m s⁻¹ accelerates at $3.0$ m s⁻² over a distance of $24$ m.

Calculate the car's final velocity.

Excellence

A stone is thrown vertically upward at $15$ m s⁻¹ (take $g = 9.8$ m s⁻² downward).

Calculate the maximum height the stone reaches, and explain the sign convention you used and why the velocity is zero at the highest point.

Constant acceleration and the equations of motion

Key ideas

Step 1 — List what you know and want

Step 2 — Final velocity (the equation with no ddd)

Step 3 — Distance (the equation with no vfv_fvf​)

Free fall

Practice question

Test yourself

Practice by grade

Constant acceleration and the equations of motion

Key ideas

Step 1 — List what you know and want

Step 2 — Final velocity (the equation with no ddd)

Step 3 — Distance (the equation with no vfv_fvf​)

Free fall

Practice question

Test yourself

Practice by grade

Step 2 — Final velocity (the equation with no $d$ )

Step 3 — Distance (the equation with no $v_f$ )

Step 2 — Final velocity (the equation with no $d$ )

Step 3 — Distance (the equation with no $v_f$ )