Question

Match Column - $\text{I}$ with Column -$\text{II}$
$\begin{array}{cc}\text{Column - I}& \text{Column - II}\\ \text{i.Velocity of a particle following}& \text{(p)}\alpha \\ \text{equation}x=u(t-8)+\alpha (t-2{)}^2& \\ \text{at}t=0& \\ \text{ii.Acceleration of a particle moving according to equation}& \text{(q)}2\alpha \\ x=u(t-8)+\alpha (t-2{)}^2\text{at}t=0& \\ \text{iii.Velocity of the particle moving}& \text{(r) u}\\ \text{according to equation}x=ut+& \\ 5\alpha t^2\ln (1+\frac{t}{2})+\alpha t^2\text{at}t=0& \\ \text{iv.Acceleration of the particle moving}& \text{(s)}u-4\alpha \\ \text{according to the equation}x=ut+& \\ 5\alpha t^2(1+\frac{t}{2})+\alpha t^2\text{at}t=0& \end{array}$

• A. $\text{i}\to \left(s\right);\text{ii}\to \left(q\right);\text{iii}\to \left(r\right);\text{iv}\to \left(q\right)$
• B. $\text{i}\to \left(s\right);\text{ii}\to \left(q\right);\text{iii}\to \left(s\right);\text{iv}\to \left(p\right)$
• C. $\text{i}\to \left(r\right);\text{ii}\to \left(p\right);\text{iii}\to \left(s\right);\text{iv}\to \left(q\right)$
• D. $\text{i}\to \left(r\right);\text{ii}\to \left(r\right);\text{iii}\to \left(r\right);\text{iv}\to \left(q\right)$

Answer 1

The correct option is A $\text{i}\to \left(s\right);\text{ii}\to \left(q\right);\text{iii}\to \left(r\right);\text{iv}\to \left(q\right)$

$\text{i}.x=u(t-8)+\alpha (t-2{)}^2$
Differentiating w.r.t t, we get
$v=u+2\alpha (t-2)\times 1$
$\Rightarrow v=u+2\alpha (t-2)$
At $t=0$
$v=u-4\alpha$
$\text{ii}.x=u(t-8)+\alpha (t-2{)}^2$
Differentiating w.r.t t we get
$v=u+2\alpha (t-2)$
Again differentiating w.r.t t we get
$a=2\alpha$
At $t=0,a=2\alpha$
$\text{iii}.x=ut+5\alpha t^2\ln (1+\frac{t}{2})+\alpha t^2$
Differentiating w.r.t $t$ , we get
$v=u+10\alpha t\ln (1+\frac{t}{2})+\frac{5.\alpha t^2.\frac{1}{2}}{1+\frac{t}{2}}+2\alpha t$
$=u+10\alpha t\ln (1+\frac{t}{2})+\frac{5\alpha t^2}{2+t}+2\alpha t$
At $t=0,$
$v=u+0+0+0=u$
$\text{iv}.x=ut+5\alpha t^2\ln (1+\frac{t}{2})+\alpha t^2$
$\Rightarrow v=u+10\alpha t\ln (1+\frac{t}{2})+\frac{5\alpha t^2}{2+t}+2at$
Differentiating w.r.t $t$, we get
$a=10\alpha .\ln (1+\frac{t}{2})+10\alpha t.\frac{\frac{1}{2}}{1+\frac{t}{2}}+\frac{10\alpha t(2+t)-5\alpha t^2}{(2+t{)}^2}+2\alpha$
At $t=0$
$a=0+0+0+2\alpha =2\alpha$