Exercises — Equations of motion (SUVAT) — derivations from calculus

List: $u=3$, $a=2$, $t=5$, want $v$, don't care about $s$. Missing letter: $s$ → use $v=u+at$. Why this one? It is the only equation with no $s$, so $s$ never has to be guessed. $v=3+(2)(5)=13\text{m/s}$

List: $s=60$, $u=4$, $v=16$, want $t$, and $a$ is not given and not wanted. Missing letter: $a$ → use $s=\frac{1}{2}(u+v)t$. Why? No $a$ appears, so we never need it. $60=\frac{1}{2}(4+16)t=10t\Rightarrow t=6\text{s}$

List: $u=30$, $v=0$, $s=90$, want $a$, time $t$ not given. Missing letter: $t$ → use $v^2=u^2+2as$. $0=30^2+2a(90)\Rightarrow 0=900+180a\Rightarrow a=-5{\text{m/s}}^2$ The minus means the acceleration points opposite to the motion — braking, exactly as expected.

List: $u=20$, $a=-9.8$, at the top $v=0$, want $s$, time $t$ not given. Missing letter: $t$ → use $v^2=u^2+2as$. $0=20^2+2(-9.8)s\Rightarrow 0=400-19.6s\Rightarrow s=\frac{400}{19.6}=20.41\text{m}$ Why $a$ negative? Up is our positive direction; gravity pulls down, so it subtracts from velocity. If you'd written $a=+9.8$ you'd get a negative height — a nonsense sign that flags the error.

List: $u=20$, $a=-9.8$, $s=0$ (back at same height), want $t$, final $v$ not wanted. Missing letter: $v$ → use $s=ut+\frac{1}{2}a{t}^2$. $0=20t+\frac{1}{2}(-9.8)t^2=20t-4.9t^2=t(20-4.9t)$ Two solutions: $t=0$ (the throw itself) and $20-4.9t=0\Rightarrow t=\frac{20}{4.9}=4.08\text{s}$. Why two answers? The equation describes the whole parabola of height vs time; it crosses $s=0$ twice — leaving and returning. We want the second crossing.

List: $u=25$, $v=0$, $t=20$, want $s$, $a$ not wanted. Missing letter: $a$ → use $s=\frac{1}{2}(u+v)t$. $s=\frac{1}{2}(25+0)(20)=\frac{1}{2}(25)(20)=250\text{m}$ This is just the area of the triangle under the $v$–$t$ graph — see Velocity-time graphs.

List: $u=5$, $a=9.8$, $s=40$, want $v$, time $t$ not given. Missing letter: $t$ → use $v^2=u^2+2as$. $v^2=5^2+2(9.8)(40)=25+784=809$ $v=\sqrt{809}=28.44\text{m/s}$ We take the positive root because the stone moves in the positive (downward) direction throughout — no turning point occurs. (The negative root would describe a stone launched upward that arrives at that speed, which is not this scenario.)

List: $u=0$ (from rest), $v=24$, $t=12$, want $a$ then $s$. (a) Straight line from rest: slope $a=\frac{v-u}{t}=\frac{24-0}{12}=2{\text{m/s}}^2$. (b) In the first 6 s: $u=0$, $a=2$, $t=6$, want $s$ → missing $v$ → $s=ut+\frac{1}{2}a{t}^2$. $s=0+\frac{1}{2}(2)(6^2)=36\text{m}$ On the graph this is the shaded triangle under the line up to $t=6$: base $6$, height $v(6)=2\times 6=12$, area $\frac{1}{2}(6)(12)=36\text{m}$. ✓ Both routes agree.

List: $u=-8$, $a=2$; part (a) wants the time when $v=0$; part (b) wants $s$ at $t=8$. (a) Rest means $v=0$. Use $v=u+at$: $0=-8+2t\Rightarrow t=4\text{s}$. The object first slows, stops at $t=4$, then reverses. (b) $s=ut+\frac{1}{2}a{t}^2=(-8)(8)+\frac{1}{2}(2)(8^2)=-64+64=0\text{m}$. Displacement is zero at $t=8$: it travelled out to a farthest point (at $t=4$, $s=-8(4)+\frac{1}{2}(2)(16)=-32+16=-16\text{m}$) then returned exactly to the start. Distance travelled is $32\text{m}$ but displacement is $0$ — the distance/displacement distinction from the parent note.

List: $u=10$; distance in the 5th second is $22$; want $a$. Distance in the $n$-th second = (distance up to $n$) minus (distance up to $n-1$). Use $s=ut+\frac{1}{2}a{t}^2$ at $t=5$ and at $t=4$: $s_5=u(5)+\frac{1}{2}a(5{)}^2=5u+\frac{25}{2}a$ $s_4=u(4)+\frac{1}{2}a(4{)}^2=4u+\frac{16}{2}a$ Now subtract term by term — the $u$-terms give $5u-4u=u$, and the $a$-terms give $\frac{1}{2}a(25-16)=\frac{9}{2}a$: $s_5-s_4=(5u-4u)+\frac{1}{2}a(25-16)=u+\frac{9}{2}a$ Set this equal to the given $22$ m and solve: $22=10+4.5a\Rightarrow 4.5a=12\Rightarrow a=\frac{12}{4.5}=2.667{\text{m/s}}^2$

List (phase 1): $u=0$, $a_1=3$, $t_1=4$, want top speed $v_1$ and distance $s_1$. Top speed $v_1=u+a_1{t}_1=0+3(4)=12\text{m/s}$. Distance: $s_1=\frac{1}{2}(u+v_1)t_1=\frac{1}{2}(0+12)(4)=24\text{m}$. List (phase 2): speed constant at $v_1=12$, so $a=0$, $t_2=6$, want $s_2$. $s_2=v_1{t}_2=12\times 6=72\text{m}$. Total: $s=24+72=96\text{m}$. Why split it? SUVAT needs $a$ constant within a phase. The acceleration changes at $t=4$s, so we treat each stretch separately and the top speed of phase 1 becomes the (constant) speed of phase 2.

List (across the window): $s=1.5$, $t=0.2$, $a=9.8$, unknown entry speed $u=v_{\text{top}}$. Use $s=ut+\frac{1}{2}a{t}^2$ with $u=v_{\text{top}}$: $1.5=v_{\text{top}}(0.2)+\frac{1}{2}(9.8)(0.2^2)=0.2v_{\text{top}}+0.196$ $0.2v_{\text{top}}=1.304\Rightarrow v_{\text{top}}=6.52\text{m/s}$ List (release to window top): $u=0$ (from rest), $v=6.52$, $a=9.8$, want $h$, time unknown. Missing letter: $t$ → use $v^2=u^2+2as$: $6.52^2=0^2+2(9.8)h\Rightarrow h=\frac{42.5104}{19.6}=2.169\text{m}$ So it was released about $2.17\text{m}$ above the window's top.

List: measure height $y$ from the ground, up positive. Both balls share $a=-9.8$.

• Ball A: $y_A=25t-4.9t^2$ (starts at $0$, moves up).
• Ball B: $y_B=40+0\cdot t-4.9t^2=40-4.9t^2$ (starts at $40$, dropped). They meet when $y_A=y_B$: $25t-4.9t^2=40-4.9t^2$ The $-4.9t^2$ terms cancel — both feel identical gravity, so their gap closes at the constant approach rate $25\text{m/s}$: $25t=40\Rightarrow t=1.6\text{s}$ Height: $y_A=25(1.6)-4.9(1.6^2)=40-12.544=27.456\text{m}$. The insight (see Figure 2): because gravity affects both equally, the relative motion is just closing a $40\text{m}$ gap at $25\text{m/s}$ — a beautifully simple hidden structure. (See how this thinking feeds Projectile motion.)

List (reaction phase): $u=20$, $a=0$, $t_r=0.8$, want $s_r$. $s_r=u{t}_r=20(0.8)=16\text{m}$. List (braking phase): $u=20$, $v=0$ (stops), $a=-6$, want $s_b$, time not wanted. Missing letter: $t$ → use $v^2=u^2+2as$: $0=20^2+2(-6)s_b\Rightarrow s_b=\frac{400}{12}=33.33\text{m}$ Total: $s=16+33.33=49.33\text{m}$. Why split at reaction? During reaction time acceleration is $0$ (constant speed); braking has a different constant $a$. Two phases, each with its own constant acceleration — SUVAT applied twice.

List: at release the packet shares the balloon's velocity, so $u=+6\text{m/s}$ (upward), not zero. Ground is $y=0$, so displacement from release is $s=-50\text{m}$; $a=-9.8$; want $t$ then $v$. Time first, using $s=ut+\frac{1}{2}a{t}^2$: $-50=6t-4.9t^2\Rightarrow 4.9t^2-6t-50=0$ $t=\frac{6\pm \sqrt{36+4(4.9)(50)}}{2(4.9)}=\frac{6\pm \sqrt{1016}}{9.8}=\frac{6\pm 31.875}{9.8}$ Take the positive time: $t=\frac{37.875}{9.8}=3.865\text{s}$ (the negative root is a fictitious "before release" time). Impact speed via $v=u+at=6+(-9.8)(3.865)=6-37.88=-31.88\text{m/s}$. The minus shows it moves downward; speed on impact is $31.88\text{m/s}$.

Write the five letters, fill in what you know, cross out the letter you don't care about, and use the equation missing it. If the motion changes its rule partway — reaction then braking, up then down, one phase then another — chop it into pieces where $a$ stays constant, solve each piece with its own List, and hand the end-of-one-piece velocity to the start of the next. That single habit — List, cross out, pick, and split at every change of rule — carries you through every problem on this page.