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 $v=u+at$


$=70\ m/s$

Therefore, velocity after $7\ seconds$ is $70\ m/s$.

Updated on: 10-Oct-2022


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