Paper 1 Kinematics Answers

Q	Ans	Brief reason
1	B	Distance $=500\mathrm{m}$, displacement $=100\mathrm{m}$ east.
2	C	Opposite velocity and acceleration directions mean speed decreases.
3	C	Crossing the time axis means $v=0$ at that instant.
4	B	Gradient of $x$-$t$ gives instantaneous velocity.
5	C	Signed area under $v$-$t$ gives displacement.
6	C	$s=\frac{1}{2}a{t}^2=25\mathrm{m}$.
7	B	$a=(0-18)/6=-3.0\mathrm{m}{\mathrm{s}}^{-2}$.
8	B	SUVAT assumes constant acceleration over the interval.
9	B	Velocity is zero momentarily; acceleration remains downward.
10	A	$v=u+at=12-15=-3.0\mathrm{m}{\mathrm{s}}^{-1}$.
11	B	Acceleration is opposite to velocity.
12	B	Increasing positive gradient means increasing positive velocity.
13	B	Horizontal line below axis means constant negative velocity.
14	A	$\Delta v=at=-12\mathrm{m}{\mathrm{s}}^{-1}$.
15	B	$s=\frac{1}{2}g{t}^2=19.6\mathrm{m}$.
16	C	At terminal velocity, drag balances weight.
17	A	Acceleration is the gradient of $v$-$t$.
18	B	Constant non-zero acceleration gives a linear $v$-$t$ graph with non-zero gradient.
19	A	Position and velocity are different quantities.
20	C	Distance is accumulated path length, never signed displacement.