Question

An object starting from rest travels $20\mathrm{m}$ in first $2\mathrm{s}$ and $160\mathrm{m}$ in next $4\mathrm{s}$. What will be the velocity after $7\mathrm{s}$ from the start?

Answer 1

Step 1:- Given data

Initial velocity $=u=0m/s$

Distance traveled $=S=20m$ in time $=t=2sec.$

Distance traveled $=160m$ in time $4sec.$
Step 2:- From the second equation of motion
$S=ut+\frac{1}{2}at²$

Where $S$is displacement, $u$is initial velocity, $a$is acceleration and $t$ is time.
$20=0\times 2+\frac{1}{2}a\times 2^2$
$20=2a$
$a=10m/s²$
Step 3:- Velocity at the end of $2sec$:
$$\begin{gathered} V=u+at \\ V=0+10\times 2 \\ V=20m/s \end{gathered}$$
Step 4:- In next $4sec$, it covers a distance of $160m$
So, from the second equation of motion,
$$\begin{gathered} 160=20\times 4+\frac{1}{2}(a\text{'})4² \\ 160=80+8a^{\text{'}} \\ 8a\text{'}=80 \\ a\text{'}=10m/s² \end{gathered}$$
It shows acceleration is uniform.
Step 5:- Velocity after $7sec$.

So, the velocity at end of $7sec$ from start:
$$\begin{gathered} V=u+at \\ V=0+10\times 7 \\ V=70m/s \end{gathered}$$
Hence, the velocity after $7sec$ from the start is $70m/s$.