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An object starting from rest travels $20\ m$ in first $2\ s$ and $160\ m$ in next $4\ s$. What will be the velocity after $7\ s$ from the start.
As given, initial velocity $u=0$, time $t=2\ s$, distance travelled $s=20\ m$
On using second equation of motion, $s=ut+\frac{1}{2}at^2$
Or $20=0+\frac{1}{2}\times a\times 2^2$
Or $20=2a$
Or $a=\frac{20}{2}$
Or $a=10\ ms^2$
after this journey, velocity obtained $v=u+at$
$=0+10\times2=20\ m/s$
For next journey of $s'=160\ m$, time $t'=4\ s$, initial velocity $u'=20\ m/s$.
Let $a'$ be the acceleration for the next 4-second journey,
So, $s'=u't'+\frac{1}{2}a'(t')^2$
Or $160=20\times4+\frac{1}{2}a'(4)^2$
Or $160=80+8a'$
Or $8a'=160-80=80$
Or $a'=\frac{80}{8}=10\ m/s^2$
Here, acceleration remains the same,
So, the final velocity after 7 seconds from the start $v=u+at$
$=0+10\times7$
$=70\ m/s$
Therefore, velocity after $7\ seconds$ is $70\ m/s$.