Kinematics: displacement, velocity, acceleration, distance – IB Maths exam questions with mark scheme

1HardNon-calculator8 marks

A particle moves in a straight line. Its velocity at time $t$ seconds, $t\ge0$, is $v=3t^2-12t+9\ \text{m s}^{-1}$. When $t=0$ the displacement of the particle from the origin is $2$ m.

(a) Find the times at which the particle is instantaneously at rest. [2]

(b) Find an expression for the displacement $s$ of the particle at time $t$. [3]

Sets $3t^2-12t+9=0$ (or $t^2-4t+3=0$)M1
$t=1$ and $t=3$A1
Integrates $v$ (attempt)M1
$s=t^3-6t^2+9t+c$A1
Uses $s(0)=2$: $s=t^3-6t^2+9t+2$A1
Identifies direction changes at $t=1,3$ and evaluates $s$ at $t=0,1,3,4$ (or splits $\int_0^4|v|\,dt$ at $t=1$ and $t=3$)M1
$s=2,\ 6,\ 2,\ 6$ respectively (or the three part-integrals $4,\,-4,\,4$)A1
Distance $=4+4+4=12$ mA1
Answer: (a) $t=1,\,3$; (b) $s=t^3-6t^2+9t+2$; (c) $12$ m
Examiner tip: Displacement from $s(4)-s(0)$ is 4 m, not 12 m; split the motion wherever $v$ changes sign.

2CoreNon-calculator6 marks

A particle moves along a straight line so that its displacement from a fixed point $O$ after $t$ seconds is $s=2t^3-15t^2+24t+5$ metres, $t\ge0$.

(a) Find an expression for the velocity $v$ of the particle at time $t$. [2]

(b) Find the acceleration of the particle when $t=3$. [2]

Attempts to differentiate $s$ (at least two terms correct)M1
$v=6t^2-30t+24$A1
Differentiates $v$: $a=12t-30$M1
$a=6\ \text{m s}^{-2}$ (ignore missing units)A1
Sets $12t-30=0$ to get $t=2.5$ and substitutes into $s$M1
$s=2.5$ m (accept $\frac52$)A1
Answer: (a) $v=6t^2-30t+24$; (b) $6\ \text{m s}^{-2}$; (c) $2.5$ m
Examiner tip: In (c) substitute $t=2.5$ into $s$, not into $v$ or $a$: the question asks for a displacement.

3CoreCalculator6 marks

A cyclist travels along a straight path. Her velocity, in metres per second, $t$ seconds after passing a marker post is $v(t)=20e^{-0.4t}-6$, for $0\le t\le 8$.

(a) Find the acceleration of the cyclist when $t=2$. [2]

(b) Find the value of $t$ at which the cyclist is instantaneously at rest. [2]

Attempts $v'(t)$ or uses GDC derivative at $t=2$: $a=-8e^{-0.4t}$M1
$a=-3.59\ \text{m s}^{-2}$ (accept $-3.5946\ldots$; accept $-8e^{-0.8}$)A1
Sets $20e^{-0.4t}-6=0$ (or finds the zero of $v$ on GDC)M1
$t=3.01$ (accept $2.5\ln\frac{10}{3}$)A1
Uses $\displaystyle\int_0^8 |v(t)|\,dt$ (or splits the integral at $t=3.01$ and adds magnitudes)M1
$33.9$ m ($33.918\ldots$)A1
Answer: (a) $-3.59\ \text{m s}^{-2}$; (b) $t=3.01$; (c) $33.9$ m
Examiner tip: $\int_0^8 v\,dt$ is only about $-0.04$ here (she ends up almost back at the post), so distance needs $\int|v|\,dt$.

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4StandardNon-calculator8 marks

A particle $P$ moves in a straight line. Its acceleration, in $\text{m s}^{-2}$, at time $t$ seconds is $a=6t-2$, $t\ge0$.

When $t=1$, the velocity of $P$ is $3\ \text{m s}^{-1}$. When $t=0$, the displacement of $P$ from a fixed point $O$ is $1$ m.

(a) Find an expression for the velocity $v$ of $P$ in terms of $t$. [3]

(b) Find the displacement of $P$ from $O$ when $t=2$. [3]

Attempts to integrate $a$M1
$v=3t^2-2t+c$A1
Uses $v(1)=3$ to get $c=2$: $v=3t^2-2t+2$A1
Integrates $v$ with a constant: $s=t^3-t^2+2t+d$M1
Uses $s(0)=1$: $d=1$A1
$s(2)=9$ mA1
Considers $3t^2-2t+2=0$ and finds discriminant $(-2)^2-4(3)(2)=-20$ (or completes the square: $3\left(t-\frac13\right)^2+\frac53$)M1
Discriminant $<0$ (or minimum value $\frac53>0$) so $v\ne0$ for all $t$, hence never at restR1
Answer: (a) $v=3t^2-2t+2$; (b) $9$ m; (c) $\Delta=-20<0$ so $v>0$ always
Examiner tip: The initial condition is at $t=1$, not $t=0$: candidates who set $c=3$ get the wrong velocity and so the wrong displacement in (b).

5StandardNon-calculator6 marks

The diagram shows the velocity–time graph for a particle moving in a straight line for $0\le t\le10$, where $v$ is in $\text{m s}^{-1}$ and $t$ is in seconds. The graph consists of straight-line segments joining $(0,0)$, $(2,6)$, $(6,6)$, $(9,-3)$ and $(10,-3)$.

(a) Write down the acceleration of the particle for $6<t<9$. [1]

(b) Find the value of $t$ at which the particle changes direction. [1]

(d) Find the displacement of the particle at $t=10$ relative to its starting position. [1]

$-3\ \text{m s}^{-2}$A1
$t=8$A1
Area above the axis: e.g. $\frac12\times2\times6+4\times6+\frac12\times2\times6=36$ (or trapezium $\frac12(8+4)\times6$)M1
Area below the axis: $\frac12\times1\times3+1\times3=4.5$A1
Total distance $=36+4.5=40.5$ mA1
$36-4.5=31.5$ m (FT their areas)A1
Answer: (a) $-3\ \text{m s}^{-2}$; (b) $t=8$; (c) $40.5$ m; (d) $31.5$ m
Examiner tip: Distance adds the area below the axis; displacement subtracts it — candidates often give 31.5 for both.

6StandardNon-calculator6 marks

A particle moves in a straight line with velocity $v=t^2-6t+8\ \text{m s}^{-1}$, for $0\le t\le5$, where $t$ is in seconds.

(a) Find an expression for the acceleration $a$ of the particle. [1]

(b) Find the values of $t$ for which the particle is slowing down. Justify your answer. [4]

$a=2t-6$A1
Finds $v=0$ at $t=2$ and $t=4$, and $a=0$ at $t=3$A1
Recognises slowing down when $v$ and $a$ have opposite signs (sign diagram or table of signs of $v$ and $a$ seen)M1
$0\le t<2$ (accept $0<t<2$)A1
$3<t<4$A1
$v(2.5)=-0.75<0$ and $a(2.5)=-1<0$: same sign so speed $|v|$ is increasingR1
Answer: (a) $a=2t-6$; (b) $0\le t<2$ and $3
Examiner tip: Negative acceleration does not mean slowing down; compare the signs of $v$ and $a$.

7HardCalculator7 marks

A particle moves along a straight line. Its velocity, in $\text{m s}^{-1}$, after $t$ seconds is $v(t)=2t\cos t$, for $0\le t\le5$. When $t=0$ the particle is at the point $A$, where its displacement from a fixed point $O$ is $1$ m.

(a) Find the maximum speed of the particle for $0\le t\le5$. [2]

(b) Find the total distance travelled by the particle for $0\le t\le5$. [2]

Considers $|v|$ (e.g. graph of $|v|$ or minimum of $v$ on GDC near $t=3.43$)M1
$6.58\ \text{m s}^{-1}$ ($6.5767\ldots$)A1
Uses $\displaystyle\int_0^5|2t\cos t|\,dt$M1
$14.1$ m ($14.110\ldots$)A1
Uses $s(5)=1+\displaystyle\int_0^5 2t\cos t\,dt$M1
$\displaystyle\int_0^5 2t\cos t\,dt=-11.02$ (seen or implied)A1
$s(5)=-10.0$ m ($-10.0219\ldots$; accept $10.0$ m on the other side of $O$)A1
Answer: (a) $6.58\ \text{m s}^{-1}$; (b) $14.1$ m; (c) $-10.0$ m
Examiner tip: Maximum speed is the largest $|v|$: the local maximum of $v$ (1.12) is not the answer.

8HardNon-calculator8 marks

Two particles $A$ and $B$ start from the same point $O$ at $t=0$ and move along the same straight line. Their velocities, in $\text{m s}^{-1}$, at time $t$ seconds are

$$v_A=3t^2-4t,\qquad v_B=2t+4,\qquad t\ge0.$$

(a) Find the value of $t$, $t>0$, at which $A$ and $B$ are at the same position. [4]

(b) Find the total distance travelled by $A$ from $t=0$ until this time. Give your answer as an exact fraction. [4]

Integrates both velocitiesM1
$s_A=t^3-2t^2$ and $s_B=t^2+4t$ (constants zero as both start at $O$)A1
Equates and simplifies: $t^3-3t^2-4t=0$, i.e. $t(t-4)(t+1)=0$M1
$t=4$ (reject $t=0$ and $t=-1$)A1
Finds $v_A=0$ at $t=\frac43$M1
$s_A\left(\frac43\right)=-\frac{32}{27}$A1
$s_A(4)=32$; distance $=\frac{32}{27}+\left(32+\frac{32}{27}\right)$M1
$\frac{928}{27}$ m (accept $34\frac{10}{27}$)A1
Answer: (a) $t=4$; (b) $\frac{928}{27}$ m
Examiner tip: $A$ moves backwards first: $s_A(4)=32$ alone misses the $2\times\frac{32}{27}$ m of back-and-forth.

9StretchNon-calculatorHL only8 marks

A particle moves in a straight line, starting from a fixed point $O$. Its velocity, in $\text{m s}^{-1}$, at time $t$ seconds is $v=te^{-t}$, $t\ge0$.

(a) Use integration by parts to show that the displacement of the particle from $O$ at time $t$ is $s=1-(t+1)e^{-t}$. [4]

(b) Hence explain why the particle never reaches a point $1$ m from $O$. [2]

Attempts integration by parts with $u=t$ and $e^{-t}$ as the part to integrate (correct choice)M1
$-te^{-t}+\displaystyle\int e^{-t}\,dt$A1
$s=-te^{-t}-e^{-t}+c$A1
Uses $s=0$ when $t=0$ to get $c=1$, giving $s=1-(t+1)e^{-t}$ (AG)A1
$(t+1)e^{-t}>0$ for all $t\ge0$, so $s<1$R1
$v=te^{-t}\ge0$, so $s$ is increasing from $s(0)=0$; hence $s\ge0$ and the particle cannot reach $s=-1$ eitherR1
$a=e^{-t}-te^{-t}=0\Rightarrow t=1$M1
$v_{\max}=e^{-1}\ \text{m s}^{-1}$ (accept $\frac1e$)A1
Answer: (a) shown; (b) $v\ge0$ so $0\le s=1-(t+1)e^{-t}<1$ for all $t$; (c) $e^{-1}\ \text{m s}^{-1}$
Examiner tip: In (a) the constant of integration matters: without $s(0)=0$ there is no '1 −' term and the A1 is lost.

10HardNon-calculator8 marks

A particle moves in a straight line, starting at the origin when $t=0$. Its velocity, in $\text{m s}^{-1}$, at time $t$ seconds is $v=3t^2-2kt$, where $k$ is a positive constant. The particle is back at the origin when $t=6$.

(a) Show that $k=6$. [3]

(b) Find the total distance travelled by the particle for $0\le t\le6$. [3]

Integrates $v$: $s=t^3-kt^2\ (+c)$M1
$c=0$ since $s=0$ at $t=0$A1
Sets $s(6)=0$: $216-36k=0\Rightarrow k=6$ (AG)A1
$v=3t(t-4)=0$ gives $t=4$M1
$s(4)=64-96=-32$A1
Distance $=32+32=64$ mA1
Considers $v$ at the turning point $t=2$ ($v=-12$) and at the endpoint $t=6$ ($v=36$)M1
Maximum speed $36\ \text{m s}^{-1}$A1
Answer: (a) shown; (b) $64$ m; (c) $36\ \text{m s}^{-1}$
Examiner tip: Speed is $|v|$: the stationary point of $v$ gives only $12$; check the endpoint $t=6$ too.

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