2D Kinematics

In AS Maths we saw how to use the SUVAT equations in one direction when acceleration is constant. We also learned about the calculus of position, velocity and acceleration in one direction when acceleration is not constant. In A2 Maths, we extend these ideas to two dimensions and apply the SUVAT equations to bodies moving under constant acceleration such as projectiles and perform calculus for those moving under variable acceleration.

Constant Acceleration

We model a projectile as a particle acting under gravity only (we ignore air resistance etc) that is given some initial velocity at a given angle to the ground. Acceleration is zero in the horizontal direction and has gravitational acceleration in the vertical direction. Since acceleration is constant in both directions, we can resolve the initial velocity vector (as we did when resolving forces) and apply the SUVAT equations in the individual directions.

Questions could ask you to find the horizontal range of the particle (the horizontal distance it reaches before hitting the ground), the greatest height it reaches (found when vertical velocity is zero) or the time of flight – see Example 1.

When all of the information is given algebraically, it can be shown that, for a particle projected from the ground:

  • Time of flight: \frac{2U\sin(\alpha)}{ g}
  • Range: \frac{U^2\sin(2\alpha)}{ g}
  • Equation of trajectory: y=x\tan(\alpha)-gx^2\frac{1+\tan^2(\alpha)}{2U^2}

Note that g is the value of acceleration here and not the vector {\bf g}. We see that the equation of trajectory is quadratic in x and so, for a particle projected from the ground, we also have that the time to greatest height is half of time of flight. Questions may ask you to derive the formulae above – see Example 2.

For particles moving in a plane with constant velocity ({\bf a}={\bf 0}), the position vector of the particle {\bf r} is given by {\bf r}={\bf r}_0+{\bf v}t where {\bf r}_0 is the initial position of the particle. For particles moving with constant acceleration ({\bf a}\ne{\bf 0}), the particle has velocity {\bf v}={\bf u}+{\bf a}t and position {\bf r}={\bf u}t+\frac{1}{2}{\bf a}t^2 noting the bold vectors and the non-bold scalars. You may recognise these as SUVAT equation for more than one dimension. Questions will likely require the use of the unit vectors {\bf i} and {\bf j} that we learned about in AS Maths – see Example 3.

Variable Acceleration

Recall from AS Maths that when acceleration is variable then velocity can be found by either differentiating position or integrating acceleration – we took a look at the motion of a particle with variable acceleration moving along a straight line. Similarly, acceleration can be found by differentiating velocity or position can be found by integrating velocity. In more than one dimension, we have

{\bf v}(t)=\frac{d{\bf r}}{dt},\hspace{15pt}{\bf a}(t)=\frac{d{\bf v}}{dt}


{\bf v}(t)=\int{\bf a}{\text dt}, \hspace{15pt}{\bf r}(t)=\int{\bf v}\text{ dt}

noting the use of vectors to show that this is now in multiple dimensions. Note that sometimes the use of a dot is shorthand for differentiation with respect to time and that questions will again require the use of the unit vectors – see Example 4.


A particle is projected from point A which is at a height of 2m above the ground with initial speed 30m/s at an angle of 35^\circ to the horizontal. The particle then moves freely under gravity until it hits the ground at point B.
a) Find the time of flight of the particle.
b) Find the greatest height reached by the particle.
c) Find the distance AB.

We consider the horizontal and vertical components of motion separately where the acceleration is 0 in the positive horizontal direction and equal to gravity (g=-9.8) in the positive vertical direction. Note that we cannot use the formulae in the notes above as this only applies to projectiles fired from the ground.

The components of the initial velocity are 30\cos(35^\circ)=24.57m/s and 30\sin(35^\circ)=17.21m/s to 2 decimal places in the horizontal and vertical directions respectively.

a) The particle reaches point B when the vertical displacement of the particle is S=-20. We know that A=-9.8 and U=17.21 in the vertical direction and we want T, so we use the SUVAT equation S=UT+\frac{1}{2}AT^2 (see more on the SUVAT equations). It follows that -20=17.2T-\frac{1}{2}\times 9.8\times T^2. Solving this quadratic equation gives either T=-0.92 or T=4.43 to 2 decimal places (see more on solving quadratic equations). The negative solution tells us how long it would take the ball to hit the ground backwards in time and so we know that the time of flight must be 4.43 seconds.
b) The greatest height of the particle is reached when the vertical velocity becomes 0. Since we know the initial vertical velocity is U=17.21m/s, the vertical acceleration is A=-9.8m/s/s, final vertical velocity V=0m/s and we want S, we use the SUVAT equation V^2=U^2+2AS. Solving for S gives S=15.11 to 2 decimal places so the greatest height of the particle is 35.11m.
c) The distance AB can be found using Pythagoras and since we know that the vertical difference between A and B is 20m, we also need to find the horizontal range of the projectile. In part a), we found the time of flight T=4.43 seconds. We also know that the horizontal acceleration is A=0 and the initial horizontal velocity is U=24.57m/s. Using again the SUVAT equation S=UT+\frac{1}{2}AT^2 (or simply S=UT for horizontal displacement of projectiles) we find the range to be S=24.57\times 4.43=108.85m. It follows that the length AB is given by \vert AB\vert =\sqrt{20^2+108.85^2}=110.67m to 2 decimal places.

A cannon is fired from the ground at an elevation angle of \alpha and speed Um/s. The cannon ball moves freely under gravity until it hits the ground again.
a) Find an expression for the time of flight, T, of the cannon ball in terms of U, \alpha and gravity g.
b) Similarly, find an expression for the range of the cannon ball.
c) If the cannon ball has moved a horizontal distance x and vertical distance y, show that y=x\tan(\alpha)-\frac{gx^2}{2U^2}\left(1+\tan^2(\alpha)\right).

Again, we consider the horizontal and vertical motion separately where, in the general case, the initial velocity is U\cos(\alpha) and U\sin(\alpha) in the horizontal and vertical components respectively.

a) For a projectile fired from ground level, it is important to note that the vertical displacement is 0 when it hits the ground again. With a vertical acceleration of -gm/s/s and initial vertical velocity U\sin(\alpha)m/s, the SUVAT equation S=UT+\frac{1}{2}AT^2 then tells us that the time when displacement is 0 is when U\sin(\alpha)T-\frac{1}{2}gT^2. Factorising gives T\left(U\sin(\alpha)-\frac{1}{2}g\right)T^2=0 with solutions T=0 (which we know at the start) or T=\frac{U\sin(\alpha)}{\frac{1}{2}g}. It follows that the time of flight is T=\frac{2U\sin(\alpha)}{g} as given.
b) The horizontal range of the particle is found using the fact that time of flight is T=\frac{2U\sin(\alpha)}{g}. Horizontal acceleration is 0 and so horizontal displacement is found simply by multiplying initial velocity by time (see Example 1c). Hence, displacement is S=U\cos(\alpha)\times \frac{2U\sin(\alpha)}{g}=\frac{2U^2\sin(\alpha)\cos(\alpha)}{g}. It follows that the range of the projectile is \frac{U^2\sin(\alpha)}{g} as given using the double angle formula 2\sin(\alpha)\cos(\alpha)=\sin(2\alpha).
c) At any point in time, the horizontal displacement (S=UT) can be expressed as x=U\cos(\alpha)\times t which can be rearranged to get t=\frac{x}{U\cos(\alpha)}. Similarly, the vertical displacement (S=UT+\frac{1}{2}AT^2) can be expressed at y=U\sin(\alpha)\times t-\frac{1}{2}gt^2. Substituting the above expression for t, we get y=U\sin(\alpha)\times \frac{x}{U\cos(\alpha)}-\frac{1}{2}g\left(\frac{x}{U\cos(\alpha)}\right)^2. It follows that y=x\tan(\alpha)-\frac{gx^2}{U^2}\left(1+\tan^2(\alpha)\right) where we have used \frac{\sin(\alpha)}{\cos(\alpha)}=\tan(\alpha) and \frac{1}{\cos^2(\alpha)}=\sec^2(\alpha)=1+\tan^2(\alpha) (see more on trigonometric identities).

a) A particle moving with constant velocity has position vectors 2{\bf i}+3{\bf j} metres and 4{\bf i}-2{\bf j} metres at times t=0 seconds and t=6 seconds respectively. Find the velocity vector of the particle in m/s and the time at which the particle is due east of the origin.
b) A particle moving with constant acceleration {\bf a}={\bf i}+2{\bf j} ms^{-2} has velocity 5{\bf i}+3{\bf j} ms^{-1} after 2.5 seconds. Find the initial velocity of the particle and the bearing on which it is moving at the instant when t=3.7 seconds.

a) Since velocity is constant, we can find the velocity vector {\bf v} by finding the change in position over change in time. The change in position is given by (4{\bf i}-2{\bf j})-(2{\bf i}+3{\bf j})=2{\bf i}-5{\bf j} and so the velocity vector is {\bf v}=\frac{1}{3}{\bf i}-\frac{5}{6}{\bf j} m/s. Using the formula above {\bf r}={\bf r}_0+{\bf v}t with {\bf r}_0=2{\bf i}+3{\bf j} we have {\bf r}=2{\bf i}+3{\bf j}+(\frac{1}{3}{\bf i}-\frac{5}{6}{\bf j})t=(2+\frac{1}{3}t){\bf i}+(3-\frac{5}{6}t){\bf j}. When the particle is due east of the origin, the y-component of its position is 0. Since the y-component of position is 3-\frac{5}{6}t, the time at which the particle is due east of the origin is when t=3\div\frac{5}{6}=\frac{18}{5}=3.6 seconds.
b) Using the formula {\bf v}={\bf u}+{\bf a}t we have 5{\bf i}+3{\bf j}={\bf u}+\left({\bf i}+2{\bf j}\right)\times 2.5. It follows that the initial velocity is {\bf u}=5{\bf i}+3{\bf j}-2.5{\bf i}-5{\bf j}=2.5{\bf i}-2{\bf j}. The velocity vector after 3.7 seconds is given by 2.5{\bf i}-2{\bf j}+\left({\bf i}+2{\bf j}\right)\times 3.7=6.2{\bf i}+5.4{\bf j}. The bearing on which the particle is moving at this time is 90-\tan^{-1}\left(\frac{5.4}{6.2}\right)=48.98^\circ to two decimal places which we give as 049^\circ.

A particle is moving in a plane with velocity at time t seconds given by {\bf v}(t)=4t{\bf i}-2t^2{\bf j} ms^{-1} for t\geq 0. At time t=0, the position vector of the particle is 3{\bf j}-{\bf i} metres.
a) Find the position vector of the particle at time t seconds.
b) Find the acceleration of the particle at time t seconds.
c) Find the position, speed and acceleration of the particle after 5 seconds.

a) The position vector of the particle is found by integrating the velocity vector: \int\left(4t{\bf i}-2t^2{\bf j}\right)dt=2t^2{\bf i}-\frac{2}{3}t^3{\bf j}+{\bf c} where {\bf c} is the constant of integration vector which is given by the initial position (we see this by setting t=0). It follows that the position vector is given by {\bf r}(t)\left(2t^2+3\right){\bf i}+\left(-\frac{2}{3}t^3-1\right){\bf j} metres.
b) The acceleration vector is found by differentiating the velocity vector: {\bf a}(t)=\dot{\bf v}(t)=4{\bf i}-4t{\bf j} m/s/s.
c) The position vector after 5 seconds is {\bf r}(5)=53{\bf i}-\frac{-253}{3}{\bf j} metres. The velocity vector after 5 seconds is given by {\bf v}(5)=20{\bf i}-50{\bf j} and so the speed is given by \sqrt{20^2+(-50)^2}=10\sqrt{29} m/s. The acceleration after 5 seconds is given by 4{\bf i}-20{\bf j}.

AS Mechanics Kinematics

AS Mechanics Kinematics