Review question

# What is the acceleration of this particle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9586

## Solution

A body is travelling along a straight line with constant acceleration $\quantity{a}{m\,s^{-2}}$. The body passes a fixed point $O$ on the line with velocity $\quantity{u}{m\,s^{-1}}$. Between $\quantity{4}{s}$ and $\quantity{5}{s}$ after passing $O$ the body travels $\quantity{10}{m}$. Between $\quantity{6}{s}$ and $\quantity{7}{s}$ after passing $O$ the body travels $\quantity{12}{m}$.

Calculate the values of $u$ and $a$.

We are given distances at various times so the most useful equation of motion is going to be $s=ut+\frac{1}{2}at^2.$

We can use the equation to find the displacements at $\quantity{5}{s}$ and $\quantity{4}{s}$. $s_5=5u+\frac{25}{2}a$ $s_4=4u+\frac{16}{2}a$ Subtracting gives us the distance travelled in this period which we are told is $10$. We can do the same for the period $\quantity{6}{s}$ to $\quantity{7}{s}$. $10=u+\frac{25-16}{2}a=u+4.5a$ $12=u+\frac{49-36}{2}a=u+6.5a$ We now can subtract these equations and solve for $a$, $-2=-2a$ $a=1$ and substitute this value into one of the above equations to find $u$. $10=u+4.5$ $u=5.5$

Alternatively, we could use a graphical approach to visually achieve the same solution. If we draw a velocity-time graph, the area underneath is the distance travelled.

From the information we are given, the difference between the two shaded areas is $\quantity{2}{m}$. In the period of time between $\quantity{4}{s}$ and $\quantity{6}{s}$ the velocity increases by $2a$. So the difference between the shaded areas is a rectangle of base $1$ and height $2a$. Since we know this difference is $2$ we have $2a\times1=2$ $a=1$

Can you also deduce the value of $u$ from this diagram?