Constant Acceleration: The First Two Equations | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Constant Acceleration: The First Two Equations

When acceleration is constant in one straight line, the velocity-time graph is straight. The first two equations of motion are simply the gradient and the area of that line.

Gradient gives v = u + at

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

The initial velocity is $u$ . After time $t$ , the final velocity is $v$ .
The change in velocity is $v-u$ , so the straight-line gradient is $a=(v-u)/t$ .
Rearranging gives $v=u+at$ . The term $at$ is the change in velocity during time $t$ .

v=u+at

Symbols

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

If $a=0$ , the equation becomes $v=u$ , as it should: zero acceleration means constant velocity.

Area gives s = ½(u + v)t

The area under the straight velocity-time line is a trapezium. Its parallel sides have lengths $u$ and $v$ , and its width is $t$ .

For constant acceleration, the average velocity is halfway between the initial and final velocities.
Average velocity is $(u+v)/2$ . Multiplying by time gives the displacement.

s=\tfrac12(u+v)t

Symbols

$s$ = displacement (m)
$u$ = initial velocity (m s⁻¹)
$v$ = final velocity (m s⁻¹)
$t$ = time interval (s)

Common mistake

The average

(u+v)/2

is safe only when acceleration is constant. A curved velocity-time graph does not have a straight-line average obtained from its endpoints.

Your turnCAIE-style · original · 2 marks

A train accelerates uniformly from 6.0 m s⁻¹ to 18 m s⁻¹ in 5.0 s. Determine its acceleration and displacement.

Show worked answer

$a=(18-6.0)/5.0=2.4\ \text{m s}^{-2}$ .
$s=\tfrac12(6.0+18)(5.0)=60\ \text{m}$ .

Answer

a=2.4\ \text{m s}^{-2},\qquad s=60\ \text{m}

Constant Acceleration: The First Two Equations

When acceleration is constant in one straight line, the velocity-time graph is straight. The first two equations of motion are simply the gradient and the area of that line.

Gradient gives v = u + at

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

The initial velocity is $u$ . After time $t$ , the final velocity is $v$ .
The change in velocity is $v-u$ , so the straight-line gradient is $a=(v-u)/t$ .
Rearranging gives $v=u+at$ . The term $at$ is the change in velocity during time $t$ .

v=u+at

Symbols

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

If $a=0$ , the equation becomes $v=u$ , as it should: zero acceleration means constant velocity.

Area gives s = ½(u + v)t

The area under the straight velocity-time line is a trapezium. Its parallel sides have lengths $u$ and $v$ , and its width is $t$ .

For constant acceleration, the average velocity is halfway between the initial and final velocities.
Average velocity is $(u+v)/2$ . Multiplying by time gives the displacement.

s=\tfrac12(u+v)t

Symbols

$s$ = displacement (m)
$u$ = initial velocity (m s⁻¹)
$v$ = final velocity (m s⁻¹)
$t$ = time interval (s)

Common mistake

The average

(u+v)/2

is safe only when acceleration is constant. A curved velocity-time graph does not have a straight-line average obtained from its endpoints.

Your turnCAIE-style · original · 2 marks

A train accelerates uniformly from 6.0 m s⁻¹ to 18 m s⁻¹ in 5.0 s. Determine its acceleration and displacement.

Show worked answer

$a=(18-6.0)/5.0=2.4\ \text{m s}^{-2}$ .
$s=\tfrac12(6.0+18)(5.0)=60\ \text{m}$ .

Answer

a=2.4\ \text{m s}^{-2},\qquad s=60\ \text{m}