Question

A body is moving with uniform acceleration starting from rest.
Match column I with Column II.
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{a. Its velocity at the end of t sec., 2t sec., 3t sec. etc are in the ratio}& \text{p.}1:2:3\\ \text{b. Distances covered in}{1}^{st}\text{t sec.,}& \text{q.}1:1:1\\ 2^{nd}\text{t sec}.,3^{rd}\text{t. sec. are in the ratio}& \\ \text{c. Distances covered in t sec., 2t sec \&}& \text{r.}1:3:5\\ \text{3t sec. are in the ratio}& \\ \text{d. Change in velocity at the end of}& \text{s.}1:4:9\\ 1^{st}\text{t.sec}.,2^{nd}\text{t.sec}.,3^{rd}\text{t.sec are in the ratio}& \end{array}$

• A. $\left(a\right)\to \left(p\right);\left(b\right)\to \left(r\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(q\right)$
• B. $\left(a\right)\to \left(p\right);\left(b\right)\to \left(q\right);\left(c\right)\to \left(r\right);\left(d\right)\to \left(s\right)$
• C. $\left(a\right)\to \left(r\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(q\right)$
• D. $\left(a\right)\to \left(q\right);\left(b\right)\to \left(r\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(q\right)$

Answer 1

The correct option is A $\left(a\right)\to \left(p\right);\left(b\right)\to \left(r\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(q\right)$

$\left(a\right)\to \left(p\right);\left(b\right)\to \left(r\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(q\right)$
(a). From 1st equation of motion, we know
$v_1=u_1+a{t}_1$
$v_1=0+at$
$=at$
Similarly,
$v_2=2at\text{and}{v}_3=3at$
$\therefore v_1:v_2:v_3=1:2:3$

(b). Distance covered in first $t$ sec is given by
$d_t=\frac{1}{2}a{t}^2$
Distance covered in second $t$ sec is given by
$d_{2t}-d_t=\frac{1}{2}a(2t{)}^2-\frac{1}{2}a(t{)}^2=\frac{3}{2}a{t}^2$
Distance covered in third t sec is given by
$d_{3t}-d_{2t}=\frac{1}{2}a(3t{)}^2-\frac{1}{2}a(2t{)}^2=\frac{5}{2}a{t}^2$
$\therefore$ Ratio of distance covered is given by $1:3:5\dots \dots$

(c). Distance covered in first $t$ sec is given by
$s_1=\frac{1}{2}a{t}^2$
Distance covered in $2t$ sec is given by
$s_2=\frac{1}{2}a\left(4t^2\right)$
Distance covered in $3t$ sec is given by
$s_3=\frac{1}{2}a\left(9t^2\right)$
$\therefore s_1:s_2:s_3=1:4:9$

d. Acceleration of particle is constant, hence change in velocity in $1^{st}\text{t s},2^{nd}\text{t s and}{3}^{rd}\text{t s}$ is same.
$\text{i.e},v_1:v_2:v_3=1:1:1$