Question

A boy starting from rest is accelerated uniformly for $30\text{s}$. If $x_1,x_2,x_3$ are the distance travelled in first $10\text{s}$, next $10\text{s}$ and last $10\text{s}$ respectively, then $x_1:x_2:x_3$ is

• A. $1:2:3$
• B. $1:1:1$
• C. $1:3:5$
• D. $1:3:9$

Answer 1

The correct option is C $1:3:5$

The distance travelled in first $10\text{seconds}$ can be calculated from the second equation of motion
$x_1=ut+\frac{1}{2}a{t}^2$
$x_1=0+\frac{1}{2}a(10{)}^2=50a$
Similarly, the distance travelled in $20\text{seconds}$ will be
$x_1+x_2=ut+\frac{1}{2}a(20{)}^2$
$x_1+x_2=200a$
$x_2=150a$
and distance travelled in $30\text{seconds}$
$x_1+x_2+x_3=0+\frac{1}{2}a(30{)}^2$
$x_1+x_2+x_3=450a$
$x_3=(450-200)a=250a$
Therefore, the ratio
$x_1:x_2:x_3=1:3:5$
Therefore, option C is correct.