A.1 Kinematics

Given: $u=0$ (starts from rest), $a=3{\text{ m s}}^{-2}$, $t=8\text{ s}$. The unknown is the final velocity $v$.

The absent variable is $s$ (displacement). Use equation $v=u+at$, which does not involve $s$.

$v=0+(3)(8)=24{\text{ m s}}^{-1}$

The motorcycle's velocity after 8 seconds is $24{\text{ m s}}^{-1}$.

We need the final velocity of an object starting from rest with uniform acceleration $3{\text{ m s}}^{-2}$ after $8\text{ s}$.

This doesn't match any standard application of $v=u+at$; $0+3\times 8=24$, not 11. Eliminated.

This appears to come from $\frac{1}{2}a{t}^2=\frac{1}{2}(3)(8^2)=96$, which gives displacement, not velocity. Eliminated.

This doubles the correct answer — perhaps from forgetting to start from rest and incorrectly doubling. Eliminated.

$v=u+at=0+3\times 8=24{\text{ m s}}^{-1}$ is correct.

Given: $u=0$ (dropped from rest), $a=9.8{\text{ m s}}^{-2}$ (downward, positive), $t=4.0\text{ s}$. Unknown: $s$ (height of cliff).

The absent variable is $v$. Use $s=ut+\frac{1}{2}a{t}^2$.

$s=(0)(4.0)+\frac{1}{2}(9.8)(4.0{)}^2=0+\frac{1}{2}(9.8)(16)=78.4\text{ m}$

The cliff is $78.4\text{ m}$ high.

Find the displacement of a free-falling object dropped from rest after $4.0$ s using $s=\frac{1}{2}g{t}^2$.

This equals $\frac{1}{2}(9.8)(2^2)$, which corresponds to $t=2\text{ s}$, not $4\text{ s}$. Eliminated.

This is $9.8\times 4=39.2$, forgetting the factor of $\frac{1}{2}$ and not squaring time. Eliminated.

This is $9.8\times 4^2=156.8$, omitting the $\frac{1}{2}$ factor. Eliminated.

$s=\frac{1}{2}(9.8)(4.0{)}^2=78.4\text{ m}$ is the correct calculation.