Kinematics (DP IB Applications & Interpretation (AI): HL): Exam Questions

A golf ball is hit from a point O on a horizontal golf course, and travels at all times in a vertical plane that passes through O.  The trajectory of the golf ball is modelled by the equation

$y=-\frac{x^2}{180}+x$

where $x$  and $y$  are respectively the horizontal and vertical displacements, in metres, of the golf ball relative to point O.  Upwards is taken to be the positive direction for the vertical displacement.

Write down an appropriate domain for the model, explaining why the domain given is suitable within the context of the question.

Sketch the graph of $y$ against $x$ , labelling any intersections with the coordinate axes.

Find the maximum height reached by the golf ball.

A ball is placed at the midpoint, X, of a horizontal tube of length 120 cm.  A player is positioned at each end of the tube and is required to pump air into the tube in order to move the ball.

The velocity-time graph for a 20 second game is shown below, where $v$  is the velocity of the ball in $\mathrm{cm}{\mathrm{s}}^{-1}$  and $t$  is the time in seconds since the start of the game.

Write down the maximum speed of the ball during the game and the time interval during which that maximum speed occurs.

There are two periods of time during which the ball is decelerating. Find the magnitude of the deceleration that the ball undergoes after it has changed direction for the first time.

Find the total distance that the ball travels in the game.

A particle moves along a straight line relative to a fixed point, P.  The motion of the particle can be modelled by the function

$s\left(t\right)=\frac{t^3}{20}-\frac{17t^2}{8}+18t-4,0\le t\le 30$

where $s$  is the horizontal displacement in metres from point P and $t$  is the time in seconds.

Write down the initial distance of the particle from point P.

Find the final displacement of the particle from P.

Find an expression, in terms of $t$ , for the velocity of the particle.

Using your answer to part (c), find the times at which the particle is stationary.

A skier performs a jump in a competition.  She leaves the ground at point Q and travels in a vertical plane through Q, landing at point R.  This can be seen in the diagram below.

The trajectory of the jump can be modelled as

$h\left(x\right)=-\frac{x}{80}\left(x-35\right)+30,0\le x<70$

where $x$  is the horizontal displacement of the skier from point Q and $h$  is the vertical displacement of the skier relative to point R.

Write down the vertical distance between points Q and R.

Show that when the skier is again at the same vertical height from which she started, then her horizontal distance from point Q is 35 m.

Find an expression for $h^{\text{'}}\left(x\right)$  and hence find the total distance in the vertical direction that the skier will travel.  Show your working.

A particle moves in a straight line with a velocity $v$  ms^-1 given by $v\left(t\right)=\sqrt{\left(t^2+2t\right)}-3$ , where $t$  is measured in seconds such that  $0\le t\le 5$.

Find the acceleration of the particle at time $t=3.7.$

.

(i)       Find the change in the particle’s displacement between the times  $t=0$  and  $t=1$ .

(ii) Explain what this change in displacement tells you about the particle’s motion between those two times.

Find the total distance travelled by the particle.

A marble is projected along a marble run that travels in a vertical plane through a fixed point O.  The marble’s vertical distance, $h$ , in cm above point O can be modelled by

$h\left(t\right)=15-10t+2t^2,0\le t\le 6$

where $t$  is the time in seconds after the marble is projected.

Write down the initial height of the marble relative to the fixed point O.

Find the time at which the marble reaches its lowest point, and find the total vertical distance that the marble has travelled up to that time.

Find the velocity and acceleration of the marble at the end of the given time period.

A particle is found to have a velocity, $v$  ms^-1, that can be expressed by the function

 $v=t^{3}cost,t\ge 0$

Find an expression for the acceleration, $a$, of the particle

Hence find the acceleration of the particle at the time $t=5.2$ .

A particle moves along a straight line such that its displacement $s$ in metres from a fixed point O is given by

$s\left(t\right)=\frac{1}{2}t-\sin 2t,\mathrm{for}0\le t\le 6$

where $t$ is the time in seconds.

Write down the number of changes of direction that the particle makes.

Find an expression for the velocity of the particle at time $t$.

Find the maximum velocity and the time(s) at which it occurs.

Considering the total distance travelled by the particle, calculate the percentage of that total distance that the particle travels in the first 2 seconds of its movement.

A particle is moving along a straight line.  The position of the particle at time  seconds, measured in metres relative to a fixed origin point, is denoted by $x\left(t\right)$ .

The particle starts at rest at the origin at time $t=0$ ,  and its motion over the next six seconds is described by the equation

$\ddot{x}\left(t\right)=\frac{4{\pi }^2}{9}\sin \left(\frac{2\pi t}{3}\right)-\frac{{\mathrm{\pi }}^2}{36}\cos \left(\frac{\pi t}{12}\right),0\le t\le 6$

Find expressions for

(i)       $\dot{x}\left(t\right)$

(ii)      $x\left(t\right)$.

Find the maximum value that each of the following quantities takes on during the first six seconds of movement, as well as the time $t$ at which those maximum values occur:

(i)      the distance of the particle from the origin

(ii)     the speed of the particle

(iii)    the magnitude of the particle’s acceleration.

Find the total distance travelled by the particle during the first six seconds of its movement.

A particle is moving along a straight line.  The position of the particle at any given time, measured in metres relative to a fixed origin point, is denoted by $x$.

It is known that the velocity, $v{\mathrm{ms}}^{-1}$, of the particle is dependent on the particle’s position, and that the velocity may be described by the function

$v\left(x\right)=-\sqrt{4-9x^2},-\frac{2}{3}\le x\le \frac{2}{3}$

Show that the acceleration, $a{\mathrm{ms}}^{-2}$, of the particle may be expressed in the form

$a=v\frac{dv}{dx}$

Hence find a function giving the acceleration of the particle in terms of $x$.

Identify the minimum and maximum values of

          (i)     the speed of the particle

          (ii)     the magnitude of the particle’s acceleration

along with the values of $x$ for which those occur.

A toy car starts from a fixed point P and moves along a horizontal race track.  The horizontal displacement, $s$  cm, of the toy car from point P can be modelled by the function

$s\left(t\right)=\frac{1}{1000}t(t^2-190t+8400),0\le t\le 140$

where $t$  is the time in seconds since leaving point P.

Sketch a graph of $s\left(t\right)$  against $t$ .

Find an expression for $a\left(t\right)$ and hence find the acceleration of the toy car at $t=85$seconds.

Find the greatest speed that the toy car reaches when travelling such that its displacement is decreasing.

The level of water, $h$ m, in an estuary relative to the mean sea level, is observed over a  24-hour period starting from midnight and is modelled by the function

$h\left(t\right)=\sin (0.262t-0.5)-3\cos (0.524t)+1$

where $t$  is the time in hours after midnight.

Find the rate of change of the height of the water level at 7 am.

Find the percentage of time within the 24-hour period that the water level remains above the mean sea level.

The scientists observing the water level in the estuary anchor a buoy in place such that its horizontal position is fixed, but it is able to move up and down vertically with the changing water level

Find the total vertical distance that the buoy moves during the 24-hour time period.

For a particle P moving in a straight line let  $\frac{dv}{dt}=\left(\sin \right(t\left)\right)\left(\cos \right(4t\left)\right)$  for  $0\le t\le 2$ ,  where $v$ is the velocity of the particle and $t$ is the elapsed time in seconds.

Sketch the graph of $\frac{dv}{dt}$ on the grid below.

Find the times at which the points of inflection would occur on a displacement-time graph representing the particle’s movement, and explain the significance of these points in the context of this question.

Hence find the values of $t$ for which the displacement-time graph would be concave down.

Two particles, ${\mathrm{P}}_1$  and ${\mathrm{P}}_2$ ,  are observed moving along a straight line.  The displacements of the particles, respectively $s_1$  and $s_2$ , in metres relative to a fixed point O can be modelled for $0\le t\le 3$  by the following functions

$s_1\left(t\right)=\frac{1}{2}\sin \left(t-0.9\right)-\cos (2t-1.8)-1$

$s_2\left(t\right)=\cos (6t-5.4)-\sin (t-0.9)+2.5$

where $t$  is the time in seconds from the start of the observation.

Find an expression for the distance between the two particles at time $t$.

Hence find

(i)       the maximum distance of the particles from one another

(ii) the time at which the maximum distance between the particles occurs.

A collision occurs between the particles during the time of observation.

Find the velocity of each of the particles 0.5 seconds before the time that they collide.

A particle starts from point X and moves in a straight line. The graph below shows its velocity,$v$ ms^-1 after $t$  seconds for $0\le t\le 8$ .

The particle has an instantaneous velocity of $0$  ms^-1 at $t=0,t=3,t=5$  and $t=8$  .

The function $s\left(t\right)$ represents the displacement of the particle from point X after seconds.

It is known that the particle travels 22 metres in the first 3 seconds.

It is also known that $s\left(3\right)=s\left(7\right)$  and ${\int }_3^{5}v\mathrm{dt}=9$ .

Find the value of $s\left(5\right)-s\left(3\right)$ .

Find the total distance travelled by the particle in the first 7 seconds.

A particle P moves along a straight line.  The velocity of the particle after $t$  seconds, $v_P{\mathrm{ms}}^{-1}$  is given by

$v_P=t^2\cos \left(\frac{\mathrm{\pi }}{4}t\right),0\le t\le 10$

Write down the first value of $t$ at which P changes its direction of motion.

Find the total distance travelled by P during the periods when its speed is increasing.

A second particle, Q, also moves along a straight line.  Its velocity after $t$  seconds, $v_Q{\mathrm{ms}}^{-1}$ is given by 

$v_Q=6t+2,0\le t\le 8$

After $k$  seconds, the total distance that Q has travelled is the same as the distance that P travels during its periods of increasing speed.

Find the value of $k$ .

A particle moves in a straight line starting from point P.  The particle is found to have a velocity, $v$   ms^-1, given by the piecewise function

$v\left(t\right)=\{\begin{array}{ll}9t-3t^2& 0\le t\le 4\\ -3t+\frac{16}{t^3}-\frac{1}{4}& 4<t\le 10\end{array}$

Find the maximum velocity reached by the particle and the time at which that maximum velocity is reached.

Find an expression for the displacement of the particle from the starting point P at time $t$ , given that the displacement of the particle from point P at the end of the time period is $-119.08$  m.

Find the total distance travelled by the particle.

A particle A moves along a horizontal straight line $L_1$ .  The displacement, $s_A$  cm, of particle A from a fixed point P on $L_1$  is given by the function

$s_A\left(t\right)=\frac{1}{2}t-2t^3{e}^{-0.3t}+24,0\le t\le 22$

where $t$  is the time in seconds from the start of the motion.

Starting at the same time, another particle, B, moves along a horizontal straight line $L_2$  which is parallel to $L_1$ .

The velocity of particle B, ${\mathrm{v}}_{\mathrm{B}}\mathrm{cm}{\mathrm{s}}^{-1}$ , at time $t$  seconds is given by

$v_B\left(t\right)=\frac{1}{2}t+12,0\le t\le 22$

Find the value(s) of for which particle A is at point P.

Find the value of $t$ at which particle A first changes direction.

Find the total distance travelled by particle A in the first 8 seconds of its motion.

The displacement, $s_B$  cm, of particle B is measured relative to a fixed point Q on $L_2$ .

Given that $s_A\left(0\right)=s_B\left(5\right)$ , find:

(i) the displacement function $s_B$  for particle B

(ii) the displacement of each particle at the time when the displacement of particle A from point P is the same as the displacement of particle B from point Q.

A particle is moving along a straight line.  The position of the particle at time $t$ seconds, measured in metres relative to a fixed origin point, is denoted by $x\left(t\right)$ .

The particle starts at the origin at time  $t=0$  with a velocity of $-3{\mathrm{ms}}^{-1}$ ,  and its motion over the next ten seconds is described by the equation

$\ddot{x}\left(t\right)=\frac{3}{32}{e}^{\frac{3t}{8}}-\frac{9{\pi }^2}{25}cos\frac{3\pi t}{5},0\le t\le 10$

Considering the total distance travelled by the particle over the ten seconds, calculate the percentage of that total distance that the particle travels in the first five seconds of its movement.

Find the greatest distance from the origin point that the particle reaches, and the time $t$  at which that greatest distance is reached.

Professor Goodwin Vundera, a kinematics researcher, has been studying a particular type of particle.  The particle is known only to move along a straight line, with its acceleration,$a\mathrm{m}{\mathrm{s}}^{-2}$ , and velocity,$v{\mathrm{ms}}^{-1}$ , both dependent on the particle’s displacement, $x$ m, with reference to a fixed origin point.

The professor has defined a new real-valued function, the ‘Vundera function’, which he believes captures all necessary information about the motion of the particle.  This Vundera function, $W$, is defined by  $W\left(x\right)=\frac{a\left(x\right)}{v\left(x\right)}$ ,  where  $a\left(x\right)$ and $v\left(x\right)$ are functions describing, respectively, the particle’s acceleration and velocity in terms of $x$.  For one of the particles studied by the professor, the associated Vundera function is found to be

$W\left(x\right)=\frac{2x}{\sqrt{225-9x^2}},a<x<b$

where $a$ and $b$ are real constants.

Given that $W$ has the largest possible domain, write down the values of $a$ and $b$.

Additionally it is known that when $x=4$, the velocity of the particle is $-2{\mathrm{ms}}^{-1}$.

By using the above information and solving an appropriate indefinite integral, find expressions for the functions $v\left(x\right)$ and $a\left(x\right)$ associated with the particle.

Explain the relationship between the particle’s speed and acceleration as $x$ varies across all the values in the domain.