Question

A uniformly accelerating body covers a distance of $10\mathrm{m}$ in the $3^{\mathrm{rd}}\mathrm{second}$ and $16\mathrm{m}$ in the $6^{\mathrm{th}}\mathrm{second}$. What is the initial velocity and acceleration?

Answer 1

Step 1: Given information

Body covers a distance of $10\mathrm{m}$ in the $3^{\mathrm{rd}}\mathrm{second}$.

$$\begin{gathered} {\mathrm{s}}_1=10\mathrm{m} \\ {\mathrm{t}}_1=3\sec \end{gathered}$$

Body covers a distance of $16\mathrm{m}$ in the $6^{\mathrm{th}}\mathrm{second}$.

$$\begin{gathered} {\mathrm{s}}_2=16\mathrm{m} \\ {\mathrm{t}}_2=6\sec \end{gathered}$$

We have to determine the initial velocity and acceleration.

Step 2: Formula

We know the formula,

${\mathrm{S}}_{\mathrm{n}}=\mathrm{u}+\frac{1}{2}\mathrm{a}(2\mathrm{n}-1)$

Here,

s is the distance.

u is the initial velocity.

a is the acceleration.

Step 3: Finding acceleration

By substituting the given values in the above formula, we get

$$\begin{gathered} 10=\mathrm{u}+\left(\frac{5\mathrm{a}}{2}\right)...\left(1\right) \\ 16=\mathrm{u}+\left(\frac{11\mathrm{a}}{2}\right)...\left(2\right) \end{gathered}$$

By solving the above equations, we get

$$\begin{gathered} 16-10=\mathrm{u}+\left(\frac{11\mathrm{a}}{2}\right)-\mathrm{u}\left(\frac{5\mathrm{a}}{2}\right) \\ 6=\frac{11\mathrm{a}-5\mathrm{a}}{2} \\ 6=\frac{6\mathrm{a}}{2} \\ 12=6\mathrm{a} \\ \mathrm{a}=\frac{12}{6} \\ \mathrm{a}=2\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

By substituting the value of a in $\left(1\right)$,

$$\begin{gathered} 10=\mathrm{u}+\left(\frac{5\times 2}{2}\right) \\ 10=\mathrm{u}+5 \\ \mathrm{u}=10-5 \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the initial velocity is $5\mathrm{m}/\mathrm{s}$ and the acceleration is $2\mathrm{m}/{\mathrm{s}}^2$.