Acceleration, velocity and displacement

Acceleration is two rungs down from displacement, so differentiate twice. Each time, the chain rule brings down another 0.5:

$$v=0.5e^{0.5t},\qquad a=0.25e^{0.5t},\qquad a(4)=0.25e^{2}\approx1.8473$$

The other options are traps. (B) is the displacement at $t=4$, (C) is the velocity, and (A) is just $e^{2}$.

Product rule on $t\ln t$, then the $-t$ cancels nicely:

$$v=\ln t+t\cdot\frac{1}{t}-1=\ln t,\qquad a=\frac{1}{t},\qquad a(3)=\frac{1}{3}\approx0.33$$

You are given velocity and asked about acceleration, so differentiate once. Watch the sign on the exponential: the derivative of $-e^{-t}$ is $+e^{-t}$.

$$a(t)=2t+e^{-t}=2$$

Solve on your calculator over $0\le t\le5$ to get $t\approx0.768$. Setting the velocity equal to 2 instead is the slip to avoid.

(a) Differentiate. $\frac{1}{\pi}$ is just a constant, so it disappears: $a(t)=3\cos t$.

(b) Integrate the velocity. This time $\frac{1}{\pi}$ becomes $\frac{t}{\pi}$:

$$x(t)=\frac{t}{\pi}-3\cos t+c$$

Substitute $x\left(\frac{\pi}{3}\right)=4$. Since $\cos\frac{\pi}{3}=\frac12$, you get $\frac{1}{3}-\frac{3}{2}+c=4$, so $c=\frac{31}{6}$.

(a) "Initial" means $t=0$: $a(0)=\pi\cos0=\pi$.

(b) Integrate, dividing by $\pi$ for the inside: $v(t)=\sin(\pi t)+c$.

The condition here is $v(1)=0.5$. It is not a value at $t=0$. Since $\sin\pi=0$, $c=0.5$. Then $v(0)=\sin0+0.5=0.5$.

(c) Integrate again, using the second condition, $x(0)=0$ ("at the origin initially"):

$$x(t)=-\frac{1}{\pi}\cos(\pi t)+0.5t+k,\qquad -\frac{1}{\pi}+k=0\ \Rightarrow\ k=\frac{1}{\pi}$$

At $t=1$, $\cos\pi=-1$, so $x(1)=\frac{1}{\pi}+0.5+\frac{1}{\pi}$.

(a) Integrate, dividing by 6 for the inside:

$$x(t)=\frac{1}{6}\sin\left(6t+\frac{\pi}{2}\right)+2t+c$$

"Initially at the origin" gives $x(0)=0$. Since $\sin\frac{\pi}{2}=1$, $\frac{1}{6}+c=0$ and $c=-\frac{1}{6}$.

(b) The object starts at the origin, so its displacement after three seconds is $x(3)-x(0)=x(3)$.

Substitute $t=3$ with your calculator in radians: $x(3)\approx5.94$.

Two integrations, two hidden conditions. "From rest" gives $v(0)=0$ and "at the origin" gives $x(0)=0$, so both constants are zero:

$$v(t)=16t-\frac{t^{3}}{3}+\frac{t^{4}}{4},\qquad x(t)=8t^{2}-\frac{t^{4}}{12}+\frac{t^{5}}{20}$$

Then $x(1.5)\approx17.958$.

Integrate, dividing by $-1$ and $-2$ for the insides, then use $x(0)=0$:

$$x(t)=-40e^{-t}+20e^{-2t}+c,\qquad -40+20+c=0\ \Rightarrow\ c=20$$

So $x(t)=20-40e^{-t}+20e^{-2t}$. It starts at 0, and the exponentials die away, so it levels off at 20. The velocity is never negative for $t\ge0$, so the graph only ever rises. That is (C). (A) is the velocity graph itself, which is a common trap.

Velocity is the rate of change of displacement, so integrate from 4 to 5:

$$\int_{4}^{5}\left(0.4t^{2}+3t\right)dt=\left[\frac{0.4t^{3}}{3}+\frac{3t^{2}}{2}\right]_{4}^{5}\approx21.63$$

"Moves in one direction" tells you the distance and the displacement are the same, so a single definite integral works. You cannot integrate $\ln(t+1)$ by hand in this course, which is why it is on Paper 2. Use your calculator:

$$\int_{0}^{4}2\ln(t+1)\,dt\approx8.09$$

Integrate twice, using $v(0)=0$ and then $x(0)=0$:

$$v(t)=-1.5\cos(2t)+1.5$$

$$x(t)=1.5t-0.75\sin(2t)$$

The sensor is where the object is at $t=3$: $x(3)\approx4.710$. It detects anything within 2 m either side, so its range is $2.710\le x\le6.710$.

Since $\cos(2t)\le1$, the velocity is never negative. The object only moves forwards, so it enters the range once and leaves once. Solve $x(t)=2.710$ and $x(t)=6.710$ on your calculator: $t\approx1.689$ and $t\approx4.592$. That is $4.592-1.689\approx2.903$ seconds in range.

Average velocity is the change in displacement divided by the time taken:

$$\frac{4}{4.592-1.689}\approx1.38$$

The full walkthrough of 2024 Paper 2, Question 18 goes through it mark by mark.

Take $A$ as the origin. The snail's position at $t=15$ is the total change in its displacement, so use a definite integral on your calculator:

$$\int_{0}^{15}1.4\ln\left(1+t^{2}\right)dt\approx76.04$$

For the ant, let $u$ be its velocity at $A$. Measure its time from $t=12$. With constant acceleration 2, its velocity is $u+2(t-12)$. Integrating, with the ant at $A$ when $t=12$, gives its position:

$$x_{\text{ant}}(t)=u(t-12)+(t-12)^{2}$$

The ant passes the snail at $t=15$, so they are in the same place:

$$3u+9=76.04$$

Solve for the constant: $u\approx22.35$. Because the ant's time was measured from when it passed $A$, the constant is its velocity there.

The full walkthrough of 2022 Paper 2, Question 17 goes through it mark by mark.

First object: differentiate, $v_{1}=t^{2}-t+k$, so its initial velocity is $k$.

Second object: integrate the acceleration, $v_{2}=4t+c$. "Towards each other with the same initial speed" is the hidden condition. Same speed but opposite direction means $v_{2}(0)=-k$, so $v_{2}=4t-k$.

Equal velocities at $t=1$: $1-1+k=4-k$, so $k=2$. Now find the next time they are equal:

$$t^{2}-t+2=4t-2\ \Rightarrow\ t^{2}-5t+4=0\ \Rightarrow\ (t-1)(t-4)=0$$

So the interval is $1\le t\le4$. Both velocities are positive throughout ($t^{2}-t+2$ has no real roots, and $4t-2>0$ for $t>\frac12$), so distance equals displacement:

$$\int_{1}^{4}\left(t^{2}-t+2\right)dt=19.5,\qquad \int_{1}^{4}(4t-2)\,dt=24$$

The difference is $24-19.5=4.5$.

The full walkthrough of 2025 Paper 1, Question 18 goes through it mark by mark.