# A body starts from rest and is uniformly accelerated for 30 sec.The distance covered in first 10 sec is x_1 , for next 10 sec is x_2 and for last 10 sec is x_3 the then the ratio of X_1:X_2:X_3=?

##### 2 Answers
Apr 11, 2017

$k = 0 , 1 , 2 , 3 , \ldots .$
$1 : 3 : 5 : 7 : \left(2 k + 1\right)$

#### Explanation:

• please look at the animation and calculate the colored areas.

$k = 0 , 1 , 1 , 3 , 4. . .$
$1 : 3 : 5 : 7 : \left(2 k + 1\right)$

Apr 12, 2017

${x}_{1} : {x}_{2} : {x}_{3} = 1 : 3 : 5$

#### Explanation:

The applicable kinematic expression is
$s = u t + \frac{1}{2} a {t}^{2}$
$s$ is distance traveled in time $t$, $u$ is initial velocity and $a$ is acceleration.

Distance traveled in first $10 s$
${x}_{1} = 0 \times 10 + \frac{1}{2} a \times {10}^{2}$
$\implies {x}_{1} = 50 a$

Distance traveled in next $10 s$
${x}_{2} = {s}_{20} - {s}_{10} = \left(0 \times 20 + \frac{1}{2} a \times {20}^{2}\right) - 50 a$
$\implies {x}_{2} = 200 a - 50 a = 150 a$

Distance traveled in last $10 s$
${x}_{3} = {s}_{30} - {s}_{20} = \left(0 \times 30 + \frac{1}{2} a \times {30}^{2}\right) - 200 a$
$\implies {x}_{3} = 450 a - 200 a = 250 a$

the required ratio is
${x}_{1} : {x}_{2} : {x}_{3} = 50 a : 150 a : 250 a$
$\implies {x}_{1} : {x}_{2} : {x}_{3} = 1 : 3 : 5$