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

The correct option is C $s_2=3s_1$

Step 1, Given data

Initial velocity = $u=0$.

Let $a$ be the constant acceleration of the particle.

Step 2, Finding the relation

From the second equation of motion we have,

$s=ut+\frac{1}{2}a{t}^2$

Therefore, for first 10 seconds the distance covered by the particle is given by,

$s_1=0+\frac{1}{2}\times a\times (10{)}^2=50a$
and for next 10 seconds the distance covered by the particle is given by,

$s_2=\text{(Total distance covered in}20sec\text{)-(Total distance covered in first}10sec\text{)}$
$\Rightarrow s_2=s_{20}-s_{10}=[0+\frac{1}{2}\times a\times (20{)}^2]-50a=150a$
$\therefore \frac{s_2}{s_1}=\frac{150a}{50a}=3$
or, $s_2=3s_1$.

Hence the relation between the distances is as shown above.