# FHSST Physics/Rectilinear Motion/Equations of Motion

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# Equations of Motion

This section is about solving problems relating to uniformly accelerated motion. We'll first introduce the variables and the equations, then we'll show you how to derive them, and after that we'll do a couple of examples.

u = starting velocity (m/s) at t = 0
v = final velocity (m/s) at time t
s = displacement (m)
t = time (s)
a = acceleration (m/s²}
 ${\displaystyle v=u+at\,}$ (5.1)
 ${\displaystyle s={\frac {(u+v)}{2}}t}$ (5.2)
 ${\displaystyle s=ut+{\frac {1}{2}}at^{2}}$ (5.3)
 ${\displaystyle v^{2}=u^{2}+2as\,}$ (5.4)

${\displaystyle \Delta x=1/2*a*(\Delta t)^{2}+u(\Delta t)}$

Make sure you can rhyme these off, they are very important! There are so many different types of questions for these equations. Basically when you are answering a question like this:

1. Find out what values you have and write them down.
2. Figure out which equation you need.
3. Write it down!!!
4. Fill in all the values you have and get the answer.

 (-1.5,0) Galileo Galilei of Pisa, Italy, was the first to determine the correct mathematical law for acceleration: the total distance covered, starting from rest, is proportional to the square of the time. He also concluded that objects retain their velocity unless a force - often friction - acts upon them, refuting the accepted Aristotelian hypothesis that objects "naturally" slow down and stop unless a force acts upon them. This principle was incorporated into Newton's laws of motion (1st law).

## Equation 5.1

By the definition of acceleration ${\displaystyle a={\frac {\Delta v}{t}}}$ where ${\displaystyle \Delta }$v is the change in velocity, i.e. ${\displaystyle \Delta v=v-u}$. Thus we have

${\displaystyle {\begin{matrix}a&=&{\frac {v-u}{t}}\\v&=&u+at\end{matrix}}}$

## Equation 5.2

In the previous section we saw that displacement can be calculated from the area between a velocity-time graph and the time-axis. For uniformly accelerated motion the most complicated velocity-time graph we can have is a straight line. Look at the graph below — it represents an object with a starting velocity of u, accelerating to a final velocity v over a total time t.

To calculate the final displacement we must calculate the area under the graph — this is just the area of the rectangle added to the area of the triangle.

${\displaystyle {\begin{matrix}Area\triangle &=&{\frac {1}{2}}b\times h\\&=&{\frac {1}{2}}t\times (v-u)\\&=&{\frac {1}{2}}vt-{\frac {1}{2}}ut\end{matrix}}}$
${\displaystyle {\begin{matrix}Area\Box &=&w\times h\\&=&t\times u\\&=&ut\end{matrix}}}$
${\displaystyle {\begin{matrix}Displacement&=&Area\Box +Area\triangle \\s&=&ut+{\frac {1}{2}}vt-{\frac {1}{2}}ut\\&=&{\frac {(u+v)}{2}}t\end{matrix}}}$

## Equation 5.3

This equation is simply derived by eliminating the final velocity v in equation 5.2. Remembering from equation 5.1 that

${\displaystyle v=u+at}$

then equation 5.2 becomes

${\displaystyle {\begin{matrix}s&=&{\frac {u+u+at}{2}}t\\&=&{\frac {2ut+at^{2}}{2}}\\&=&ut+{\frac {1}{2}}at^{2}\end{matrix}}}$

## Equation 5.4

This equation is just derived by eliminating the time variable in the above equation. From Equation 5.1 we know

${\displaystyle t={\frac {v-u}{a}}}$

Substituting this into Equation 5.3 gives

 ${\displaystyle {\begin{matrix}s&=&u({\frac {v-u}{a}})+{\frac {1}{2}}a({\frac {v-u}{a}})^{2}\\&=&{\frac {uv}{a}}-{\frac {u^{2}}{a}}+{\frac {1}{2}}a({\frac {v^{2}-2uv+u^{2}}{a^{2}}})\\&=&{\frac {uv}{a}}-{\frac {u^{2}}{a}}+{\frac {v^{2}}{2a}}-{\frac {uv}{a}}+{\frac {u^{2}}{2a}}\\2as&=&-2u^{2}+v^{2}+u^{2}\\v^{2}&=&u^{2}+2as\end{matrix}}}$ (5.5)

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.

## Worked Example 28

Question: A racing car has an initial velocity of 100 m/s and it covers a displacement of 725 m in 10 s. Find its acceleration.

Step 1 :

We are given the quantities u, s and t — all in the correct units. We need to find a.

Step 2 :

We can use equation 5.3 ${\displaystyle s=ut+{\frac {1}{2}}at^{2}}$

Step 3 :

Rearranging equation 5.3 we have ${\displaystyle a={\frac {2(s-ut)}{t^{2}}}}$

Substituting in the values of the known quantities this becomes

${\displaystyle {\begin{matrix}a&=&{\frac {2(725{\mbox{ m}}-100{\frac {\mbox{m}}{\mbox{s}}}\cdot 10{\mbox{ s}})}{10^{2}{\mbox{ s}}^{2}}}\\&=&{\frac {2(-275{\mbox{ m}})}{100{\mbox{ s}}^{2}}}\\&=&-5.5{\frac {\mbox{m}}{{\mbox{s}}^{2}}}\end{matrix}}}$

The racing car is accelerating at -5.5 m/s², or we could say it is decelerating at 5.5 m/s². this ans is wrong

its better to use a=v-u/t then a=(725-100)/10s you will get a=625/10 which will be equal to 62.5m/s2

## Worked Example 29

Question: An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s. Calculate

• its acceleration
• its final velocity
• at what time the object had covered half the total distance
• what distance the object had covered in half the total time.

Step 1 : We are given the quantities u, s and t in the correct units.

Step 2 : To calculate the acceleration we can use equation [#eq:eq3 5.3]. Rearranging it we have:

${\displaystyle a={\frac {2(s-ut)}{t^{2}}}}$

Substituting in the values of the known quantities this becomes

${\displaystyle {\begin{matrix}a&=&{\frac {2(64m-0{\frac {m}{s}}4s}{4^{2}s^{2}}}\\&=&{\frac {128m}{16s^{2}}}\\&=&8{\frac {m}{s^{2}}}\end{matrix}}}$

Step 3 : To calculate its final velocity we can use equation 5.1 - remember we now also know the acceleration of the object.

${\displaystyle {\begin{matrix}v&=&u+at\\&=&0{\frac {m}{s}}+(8{\frac {m}{s^{2}}})(4s)\\&=&32{\frac {m}{s}}\end{matrix}}}$

Step 4 : The time at which the object had covered half the total distance. Half the distance is 32m. Here we have the quantities s, u and a so we first use equation 5.4 to calculate the velocity at this distance:

${\displaystyle {\begin{matrix}v^{2}&=&u^{2}+2as\\&=&(0m)^{2}+2(8m/s^{2})(32m)\\&=&512m^{2}/s^{2}\\v&=&22.6m/s\end{matrix}}}$

Now we can use equation 5.2 to calculate the time:

${\displaystyle {\begin{matrix}t&=&{\frac {2s}{u+v}}\\&=&{\frac {(2)(32m)}{0m/s+22.6m/s}}\\&=&2.8s\end{matrix}}}$

Step 5 : The distance the object had covered in half the time. Half the time is 2s. Thus we have u, a and t - all in the correct units. We can use equation 5.3 to get the distance:

${\displaystyle {\begin{matrix}s&=&ut+{\frac {1}{2}}at^{2}\\&=&(0m/s)(2s)+{\frac {1}{2}}(8{\frac {m}{s^{2}}})(2s)^{2}\\&=&16m\end{matrix}}}$

## Worked Example 30

Question: A ball is thrown vertically upwards with a velocity of 10 m/s from the balcony of a tall building. The balcony is 15 m above the ground and gravitational accleration is 10 m/s2. Find a) the time required for the ball to hit the ground, and b) the velocity with which it hits the ground.

Step 1 : In this case it often helps to make the problem easier to understand if we draw ourselves a picture like the one below:

First the ball goes upwards with gravitational acceleration slowing it until it reaches its highest point — here its speed is 0 m/s — then it begins descending with gravitational acceleration causing it to increase its speed on the way down. We can separate the motion into 2 stages:

Stage 1 - the upward motion of the ball

Stage 2 - the downward motion of the ball.

We'll choose the upward direction as positive - this means that gravitation acceleraton is negative - and we'll begin by solving for all the variables of Stage 1. So far we have these quantities:

${\displaystyle {\begin{matrix}u_{1}&=&10m/s\\v_{1}&=&0m/s\\a_{1}&=&-10m/s^{2}\\t_{1}&=&?\\s_{1}&=&?\end{matrix}}}$

Using equation 5.1 to find t1:

${\displaystyle {\begin{matrix}v_{1}&=&u_{1}+a_{1}t_{1}\\t_{1}&=&{\frac {v_{1}-u_{1}}{a_{1}}}\\&=&{\frac {0m/s-10m/s}{-10m/s^{2}}}\\&=&1s\end{matrix}}}$

We can find s1 by using equation 5.4

${\displaystyle {\begin{matrix}v_{1}^{2}&=&u_{1}^{2}+2a_{1}s_{1}\\s_{1}&=&{\frac {v_{1}^{2}-u_{1}^{2}}{2a}}\\&=&{\frac {(0m/s)^{2}-(10m/s)^{2}}{2(-10m/s^{2})}}\\&=&5m\end{matrix}}}$

For Stage 2 we have the following quantities:

${\displaystyle {\begin{matrix}u_{2}&=&0m/s\\v_{2}&=&?\\a_{2}&=&-10m/s^{2}\\t_{2}&=&?\\s_{2}&=&-15m-5m=20m\end{matrix}}}$

We can determine the final velocity v2 using equation 5.4:

${\displaystyle {\begin{matrix}v_{2}^{2}&=&u_{2}^{2}+2a_{2}s_{2}\\&=&(0m/s)^{2}+2(-10m/s^{2})(-20m)\\&=&400(m/s)^{2}\\v_{2}&=&20m/s{\mbox{ downwards}}\end{matrix}}}$

Now we can determine the time for Stage 2, t2, from equation 5.1:

${\displaystyle {\begin{matrix}v_{2}&=&u_{2}+a_{2}t_{2}\\t_{2}&=&{\frac {v_{2}-u_{2}}{a_{2}}}\\&=&{\frac {-20m/s-0m/s}{-10m/s^{2}}}\\&=&2s\end{matrix}}}$

Finally,

a) the time required for the stone to hit the ground is

${\displaystyle t=t_{1}+t_{2}=1{\mbox{ s}}+2{\mbox{ s}}=3{\mbox{ s}}}$

b) the velocity with which it hits the ground is just

${\displaystyle v_{2}=-20\ {\mbox{m}}/{\mbox{s}}}$

applying <br\> ${\displaystyle {\vec {s}}(t)={\vec {S}}_{i}+{\vec {U}}_{i}\ (t-T_{i})+{\frac {\vec {A}}{2}}(t-T_{i})^{2}}$<br\> ${\displaystyle 0=15{\hat {j}}+10{\hat {j}}(T_{total}-0)+{\frac {-10{\hat {j}}}{2}}(T_{total}-0)^{2}}$<br\> ${\displaystyle T_{total}^{2}-2\ T_{total}-3=0}$<br\> solving we will get <br\> ${\displaystyle T_{total}=3,-1}$<br\> so time to reach is 3 second , now applying <br\> ${\displaystyle {\vec {v}}(t)={\vec {U}}_{i}+{\vec {A}}(t-T_{i})}$<br\> ${\displaystyle {\vec {v}}(3)=10{\hat {j}}+(-10{\hat {j}})(3-0)}$<br\> ${\displaystyle {\vec {v}}(3)=-20{\hat {j}}}$<br\>

These questions do not have the working out in them, but they are all done in the manner described on the previous page.

Question: A car starts off at 10 m/s and accelerates at 1 m/s2 for 10 seconds. What is its final velocity?