From Mechanics

Theory

Exercises

Key Points

Vector Forms of the Constant Acceleration Equations

\displaystyle \begin{align} & \mathbf{v}=\mathbf{u}+\mathbf{a}t \\ & \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} \\ & \mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}} \\ \end{align}

Note: The first two equations are the basic equations, relating displacement, velocity and acceleration. The third equation is derived from the first two.

The position vector \displaystyle \mathbf{r} is the vector equivalent of the displacement \displaystyle s in the scalar equations.

In the above \displaystyle {{\mathbf{r}}_{0}} is the position vector at time \displaystyle t=0. This is because it may not be practical to assume the initial position is at the origin when using vectors to solve constant acceleration problems.

Distance is the magnitude of the displacement vector between two point.

Speed is the magnitude of velocity.

Example 8.1

During a 20 second period the velocity of a motor boat changes from (8 \displaystyle \mathbf{i} + 12 \displaystyle \mathbf{j} ) \displaystyle \text{m}{{\text{s}}^{-1}} to
(6 \displaystyle \mathbf{i} - 5 \displaystyle \mathbf{j} ) \displaystyle \text{m}{{\text{s}}^{-1}} , where \displaystyle \mathbf{i} and \displaystyle \mathbf{j} are perpendicular unit vectors. During the 20 seconds the acceleration of the boat is constant. Find the acceleration of the boat.

Solution

In this case \displaystyle \mathbf{u}=8\mathbf{i}+12\mathbf{j}, \displaystyle \mathbf{v}=6\mathbf{i}-5\mathbf{j} and \displaystyle t=20 . These can be substituted into the equation \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t to find \displaystyle \mathbf{a} .

\displaystyle \begin{align} & 6\mathbf{i}-5\mathbf{j}=8\mathbf{i}+12\mathbf{j}+20\mathbf{a} \\ & -2\mathbf{i}-7\mathbf{j}=20\mathbf{a} \\ & \mathbf{a}=\frac{\smash{-2\mathbf{i}-7\mathbf{j}}}{20} =-0\textrm{.}1\mathbf{i}-0\textrm{.}35\mathbf{j} \ \text{m}{{\text{s}}^{-2}} \\ \end{align}

Example 8.2

An object is initially at the origin, where it has velocity (8 \displaystyle \mathbf{i} + 5 \displaystyle \mathbf{j} ) \displaystyle \text{m}{{\text{s}}^{-1}}. As it moves it has an acceleration of (0.2 \displaystyle \mathbf{i} + 0.8 \displaystyle \mathbf{j} ) \displaystyle \text{m}{{\text{s}}^{-2}}, where \displaystyle \mathbf{i} and \displaystyle \mathbf{j} are perpendicular unit vectors. Find the position vector and velocity at time \displaystyle t seconds.

Solution

In this case \displaystyle \mathbf{u}=8\mathbf{i}+5\mathbf{j} , \displaystyle \mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}8\mathbf{j} and since the particle starts at the origin \displaystyle {{\mathbf{r}}_{0}}=0\mathbf{i}+0\mathbf{j}. Using the equation \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{ \ 2}}+{{\mathbf{r}}_{0}} allows us to obtain an expression for the position vector at any time.

\displaystyle \begin{align} & \mathbf{r}=(8\mathbf{i}+5\mathbf{j})t+\frac{1}{2}(0\textrm{.}2\mathbf{i}+0\textrm{.}8\mathbf{j}){{t}^{\ 2}}+(0\mathbf{i}+0\mathbf{j}) \\ & =8t\mathbf{i}+5t\mathbf{j}+0\textrm{.}1{{t}^{\ 2}}\mathbf{i}+0\textrm{.}4{{t}^{\ 2}}\mathbf{j} \\ & =\left( 8t+0\textrm{.}1{{t}^{\ 2}} \right)\mathbf{i}+\left( 5t+0\textrm{.}4{{t}^{\ 2}} \right)\mathbf{j} \ \text{m} \end{align}

Then using \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t an expression can be found for the velocity at any time.

\displaystyle \begin{align} & \mathbf{v}=(8\mathbf{i}+5\mathbf{j})+(0\textrm{.}2\mathbf{i}+0\textrm{.}8\mathbf{j})t \\ & =8\mathbf{i}+5\mathbf{j}+0\textrm{.}2t\mathbf{i}+0\textrm{.}8t\mathbf{j} \\ & =(8+0\textrm{.}2t)\mathbf{i}+(5+0\textrm{.}8t)\mathbf{j} \ \text{m}{{\text{s}}^{-1}} \end{align}

Example 8.3

A ball is set into motion with initial velocity (8 \displaystyle \mathbf{i} + \displaystyle \mathbf{j} ) \displaystyle \text{m}{{\text{s}}^{-1}} from the point with position vector 1.5 \displaystyle \mathbf{j} m . As it moves it has acceleration -10 \displaystyle \mathbf{j} \ \displaystyle \text{m}{{\text{s}}^{-2}}. The unit vectors \displaystyle \mathbf{i} and \displaystyle \mathbf{j} are directed horizontally and vertically respectively.

a) Find the position of the ball at time \displaystyle t seconds.

b) Find the position of the ball when it hits the ground, that is when the vertical component of its position is equal to zero.

Solution

a) In this case we have \displaystyle \mathbf{u}=8\mathbf{i}+2\mathbf{j} , \displaystyle \mathbf{a}=-10\mathbf{j} and \displaystyle {{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{j} . Substituting these into the equation \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} gives the position vector of the ball at time \displaystyle t as:

\displaystyle \begin{align} & \mathbf{r}=(8\mathbf{i}+2\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}}+1\textrm{.}5\mathbf{j} \\ & =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\ & =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m} \end{align}

b) The ball hits the ground when the \displaystyle \mathbf{j} component of the position vector is zero. That is when:

\displaystyle \begin{align} & 1\textrm{.}5+2t-5{{t}^{\ 2}}=0 \\ & 5{{t}^{\ 2}}-2t-1\textrm{.}5=0 \\ & t=\frac{2\pm \sqrt{{{2}^{2}}-4\times 5\times (-1\textrm{.}5)}}{2\times 5} \\ & t=\frac{2\pm \sqrt{34}}{10} \\ & t=-0\textrm{.}383 \ \text{s} \ \text{ or } \ t=0\textrm{.}783 \ \text{s}\\ \end{align}

As the value of \displaystyle t must be positive we select \displaystyle t=0\textrm{.}783 , so that the ball hits the ground after it has been moving for 2 seconds.

The position vector of the ball when it hits the ground can be calculated using this value of \displaystyle t and the result \displaystyle \mathbf{r}=(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} , which gives:

\displaystyle \begin{align} & \mathbf{r}=(8\times 0\textrm{.}783)\mathbf{i}+(1\textrm{.}5+2\times 0\textrm{.}783-5\times {{0\textrm{.}783}^{2}})\mathbf{j} \\ & =6\textrm{.}264\mathbf{i}+0\mathbf{j} \\ & =6\textrm{.}264\mathbf{i} \ \text{m} \end{align}

The ball travels a horizontal distance of 6\textrm{.}26 metres (correct to 3 significant figures).

Example 8.4

A boat moves, with constant acceleration from the origin, so that at time \displaystyle t its velocity is given by

\displaystyle \mathbf{v}=2(7-t)\mathbf{i}+4(3-t)\mathbf{j} , where \displaystyle \mathbf{i} and \displaystyle \mathbf{j} are unit vectors directed east and north respectively.

a) Find the time when the boat is heading due south.

b) By writing the velocity in the form \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t , find

(i) initial velocity of the boat

(ii) the acceleration of the boat.

c) Find the distance between the origin and the boat when \displaystyle t = 10 \ \text{s}.

Solution

a) When the boat is heading due south, the \displaystyle \mathbf{i} component of the velocity will be zero

\displaystyle \begin{align} & 2(7-t)=0 \\ & t=7 \ \text{s}\\ \end{align}

b) We have \displaystyle \mathbf{v}=2(7-t)\mathbf{i}+4(3-t)\mathbf{j} and can rewrite this in the form:

\displaystyle \mathbf{v}=(14\mathbf{i}+12\mathbf{j})+(-2\mathbf{i}-4\mathbf{j})t

This can then be compared with the formula \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t .

(i) Hence \displaystyle \mathbf{u}=14\mathbf{i}+12\mathbf{j} \ \text{m}{{\text{s}}^{-1}} .

(ii) Also \displaystyle \mathbf{a}=-2\mathbf{i}-4\mathbf{j} \ \text{m}{{\text{s}}^{-2}}

c) When \displaystyle t=10 \ \text{s} the position vector of the boat is given by:

\displaystyle \begin{align} & \mathbf{r}=(14\mathbf{i}+12\mathbf{j})\times 10+\frac{1}{2}(-2\mathbf{i}-4\mathbf{j})\times {{10}^{2}} \\ & =140\mathbf{i}+120\mathbf{j}-100\mathbf{i}-200\mathbf{j} \\ & =40\mathbf{i}-80\mathbf{j} \ \text{m} \end{align}

The distance can now be found using Pythagoras’ Theorem.

\displaystyle d=\sqrt{{{40}^{2}}+{{80}^{2}}}=\sqrt{8000}=89\textrm{.}4\text{ m (to 3sf)}