WORKED EXAMPLES

4) The rocket shown in Figure 2.12 lifts off from rest with an acceleration of $20\,m\,{s^{ - 2}}$. Calculate its velocity after $50 s$.
Step 1: What we know:
$\begin{array}{l}
u = 0\,m\,{s^{ - 1}}\\
a = 20\,m\,{s^{ - 2}}\\
t = 50s
\end{array}$
and what we want to know: $v = ?$
Step 2: The equation linking u, a, t and v is equation 1:
$v = u + at$
Substituting gives:
$v = 0 + (20 \times 50)$
Step 3: Calculation then gives:
$v = 1000\,m\,{s^{ - 1}}$
So the rocket will be travelling at $1000\,m\,{s^{ - 1}}$ after $50 s$. This makes sense, since its velocity increases by $20\,m\,{s^{ - 1}}$ every second, for $50 s$.
You could use the same equation to work out how long the rocket would take to reach a velocity of $2000\,m\,{s^{ - 1}}$, or the acceleration it must have to reach a speed of $1000\,m\,{s^{ - 1}}$ in $40 s$ and so on.

5) The car shown in Figure 2.13 is travelling along a straight road at $8.0\,m\,{s^{ - 1}}$. It accelerates at $1.0\,m\,{s^{ - 2}}$ for a distance of $18 m$. How fast is it then travelling?

In this case, we will have to use a different equation, because we know the distance during which the car accelerates, not the time.
Step 1: What we know:
$\begin{array}{l}
u = 8.0\,m\,{s^{ - 1}}\\
u = 8.0\,m\,{s^{ - 1}}\\
s = 18m\,
\end{array}$
and what we want to know: $v = ?$
Step 2: The equation we need is equation 4:
${v^2} = {u^2} + 2as$
Substituting gives:
${v^2} = {8.0^2} + (2 \times 1.0 \times 18)$
Step 3: Calculation then gives:
$\begin{array}{l}
{v^2} = 64 + 36 = 100\,{m^2}\,{s^{ - 2}}\\
v = 10\,m\,{s^{ - 1}}
\end{array}$
So the car will be travelling at $10\,m\,{s^{ - 1}}$ when it stops accelerating.
(You may find it easier to carry out these calculations without including the units of quantities when you substitute in the equation. However, including the units can help to ensure that you end up with the correct units for the final answer.)

6) A train (Figure 2.14) travelling at $20\,m\,{s^{ - 1}}$ accelerates at $0.50\,m\,{s^{ - 2}}$ for $30 s$. Calculate the distance travelled by the train in this time.

Step 1: What we know:
$\begin{array}{l}
u = 20\,m\,{s^{ - 1}}\\
t = 30s\\
a = 0.50\,m\,{s^{ - 2}}
\end{array}$
and what we want to know: $s = ?$
Step 2: The equation we need is equation 3:
$s = ut + \frac{1}{2}a{t^2}$
Substituting gives:
$s = (20 \times 30) + \frac{1}{2} \times 0.5 \times {(30)^2}$
Step 3: Calculation then gives:
$s = 600 + 225 = 825 m$
So the train will travel $825 m$ while it is accelerating.

7) The cyclist in Figure 2.15 is travelling at $15\,m\,{s^{ - 1}}$. She brakes so that she doesn’t collide with the wall. Calculate the magnitude of her deceleration.

This example shows that it is sometimes necessary to rearrange an equation, to make the unknown quantity its subject. It is easiest to do this before substituting in the values.
Step 1: What we know:
$\begin{array}{l}
u = 15\,m\,{s^{ - 1}}\\
v = 0\,m\,{s^{ - 1}}\\
s = 18m
\end{array}$
and what we want to know: $a = ?$
Step 2: The equation we need is equation 4:
${v^2} = {u^2} + 2as$
Rearranging gives:
$\begin{array}{l}
a = \frac{{{v^2} - {u^2}}}{{2s}}\\
a = \frac{{{0^2} - {{15}^2}}}{{2 \times 18}}\\
 = \frac{{ - 225}}{{36}}
\end{array}$
Step 3: Calculation then gives:
$a =  - 6.25\,m\,s{\,^{ - 2}}\, \approx \, - 6.3\,m\,{s^{ - 2}}$
So the cyclist will have to brake hard to achieve a deceleration of magnitude $6.3\,m\,{s^{ - 2}}$. The minus sign shows that her acceleration is negative; in other words, a deceleration.