Question

A body starts from rest and is uniformly accelerated for $30\text{s}$. The distance travelled in the first $10\text{s}$ is $x_1$, next $10\text{s}$ is $x_2$ and the last $10\text{s}$ is $x_3$. Then $x_1:x_2:x_3$ is the same as

• A. $1:2:4$
• B. $1:2:5$
• C. $1:3:5$
• D. $1:3:9$

Answer 1

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



Let acceleration of body be $a$
$x_1=\frac{1}{2}a(10{)}^2=50a$

the distance travelled in $20s$ is $x_1+x_2$
$x_1+x_2=\frac{1}{2}a(20{)}^2$
$x_2=\frac{1}{2}a(20{)}^2-\frac{1}{2}a(10{)}^2=150a$
the distance travelled in $30s$ is $x_1+x_2+x_3$
$x_1+x_2+x_3=\frac{1}{2}a(30{)}^2$
$x_3=\frac{1}{2}a(30{)}^2-\frac{1}{2}a(20{)}^2=250a$
$x_1:x_2:x_3=1:3:5$