Completing the SUVAT Equations | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Completing the SUVAT Equations

The same straight velocity-time line produces two more useful forms. One removes final velocity. The other removes time. Neither introduces new physics.

Rectangle plus triangle

Split the area below the line into a rectangle of height $u$ and a triangle whose height is the velocity change $v-u=at$ .

For constant acceleration, the gradient gives a and the area gives s. The equations of motion come from these two facts.

Rectangle area: $ut$ .
Triangle area: $\tfrac12\times t\times at=\tfrac12at^2$ .
Adding the areas gives the total displacement.

s=ut+\tfrac12at^2

Symbols

$s$ = displacement (m)
$u$ = initial velocity (m s⁻¹)
$a$ = constant acceleration (m s⁻²)
$t$ = time interval (s)

Removing time

Use this form when the question gives no time and does not ask for time. Eliminate $t$ from the two earlier equations.

From $v=u+at$ , write $t=(v-u)/a$ .
Substitute into $s=\tfrac12(u+v)t$ .
$s=\dfrac{(u+v)(v-u)}{2a}$
Use $(u+v)(v-u)=v^2-u^2$ and rearrange.

v^2=u^2+2as

Symbols

$u$ = initial velocity (m s⁻¹)
$v$ = final velocity (m s⁻¹)
$a$ = constant acceleration (m s⁻²)
$s$ = displacement (m)

Quantity not used	Choose this relation
s	v = u + at
a	s = ½(u + v)t
v	s = ut + ½at²
t	v² = u² + 2as

Common mistake

These equations need constant acceleration in a straight line. If acceleration changes, no single value of

a

represents the whole interval; use the graph or another supplied model instead.

Your turnAdapted from 9702/12/O/N/25 Q5 · 1 mark

A car has initial velocity $u$ , constant acceleration $a$ , displacement $d$ and final velocity $v$ . Write $u$ in terms of the other quantities. (9702/12/O/N/25 Q5)

Show worked answer

$v^2=u^2+2ad$ .
$u^2=v^2-2ad$ .
Speed is a magnitude, so take the positive root.

Answer

u=\sqrt{v^2-2ad}

Completing the SUVAT Equations

The same straight velocity-time line produces two more useful forms. One removes final velocity. The other removes time. Neither introduces new physics.

Rectangle plus triangle

Split the area below the line into a rectangle of height $u$ and a triangle whose height is the velocity change $v-u=at$ .

For constant acceleration, the gradient gives a and the area gives s. The equations of motion come from these two facts.

Rectangle area: $ut$ .
Triangle area: $\tfrac12\times t\times at=\tfrac12at^2$ .
Adding the areas gives the total displacement.

s=ut+\tfrac12at^2

Symbols

$s$ = displacement (m)
$u$ = initial velocity (m s⁻¹)
$a$ = constant acceleration (m s⁻²)
$t$ = time interval (s)

Removing time

Use this form when the question gives no time and does not ask for time. Eliminate $t$ from the two earlier equations.

From $v=u+at$ , write $t=(v-u)/a$ .
Substitute into $s=\tfrac12(u+v)t$ .
$s=\dfrac{(u+v)(v-u)}{2a}$
Use $(u+v)(v-u)=v^2-u^2$ and rearrange.

v^2=u^2+2as

Symbols

$u$ = initial velocity (m s⁻¹)
$v$ = final velocity (m s⁻¹)
$a$ = constant acceleration (m s⁻²)
$s$ = displacement (m)

Quantity not used	Choose this relation
s	v = u + at
a	s = ½(u + v)t
v	s = ut + ½at²
t	v² = u² + 2as

Common mistake

These equations need constant acceleration in a straight line. If acceleration changes, no single value of

a

represents the whole interval; use the graph or another supplied model instead.

Your turnAdapted from 9702/12/O/N/25 Q5 · 1 mark

A car has initial velocity $u$ , constant acceleration $a$ , displacement $d$ and final velocity $v$ . Write $u$ in terms of the other quantities. (9702/12/O/N/25 Q5)

Show worked answer

$v^2=u^2+2ad$ .
$u^2=v^2-2ad$ .
Speed is a magnitude, so take the positive root.

Answer

u=\sqrt{v^2-2ad}