Question

A particle experiences constant acceleration for $20$ seconds after starting from rest. If it travels a distance $s_1$ in the first $10$ seconds and distance $s_2$ in the next 10 seconds, then

• A. $s_2=s_1$
• B. $s_2=2s_1$
• C. $s_2=3s_1$
• D. $s_2=4s_1$

Answer 1

Answer: (C)

$s_2=3s_1$

Let $a$ be the constant acceleration of the particle. Then$s=ut+\frac{1}{2}at{}^2$ or $s_1=0+\frac{1}{2}\times a\times \left(10\right){}^2=50a$and $s_2=[0+\frac{1}{2}a{\left(20\right)}^2]-50a=150a$

$\therefore s_2=3s_1$

Alternatively:Let $a$ be constant acceleration and

$s=ut+\frac{1}{2}at{}^2$, then $s_1=0+\frac{1}{2}\times a\times 100=50a$

Velocity after 10 sec. is $v=0+10a$

So, $s_2=10a\times 10+\frac{1}{2}a\times 100=150a\Rightarrow s_2=3s_1$

Let $a$ be constant acceleration, using $s=ut+\frac{1}{2}at{}^2$

so distance coreved in first 10 seconds $s_1=0+\frac{1}{2}\times a\times 100=50a$

Velocity after 10 sec. is $v=0+10a$

So, distance covered in next 10 seconds $s_2=10a\times 10+\frac{1}{2}a\times 100=150a\Rightarrow s_2=3s_1$