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51.

Find the following integrals: (i)`int_(1)^(9)(dx)/(sqrt(x))" "` (ii) `int_(1)^(3)12x^(3)dx " "` (iii) `int_(4)^(16) sqrt(t).dt`

Answer» Correct Answer - (i) `4` ,(ii) `240` ,(iii)`(112)/(3)`
(i) `underset(1)overset(9)int(dx)/(sqrt(x))=((sqrt(x))/(-1/2+1))_(1)^(4)=2(sqrt(9-sqrt(1)))=4`
(ii)`underset(1)overset(3)int12x^(3)dx=12underset(1)overset(3)intx^(3)dx=12((x^(4))/(4))_(1)^(3)=240`
(iii)`underset(4)overset(16)intsqrt(t)dt=(2)/(3)(t^(3//2))_(4)^(16)=(112)/(3)`
Discussion
52.

Find value of `sqrt(101)`

Answer» Correct Answer - `10.05`
`sqrt(101)=(100+1)^(1//2)`
`10(1+(1)/(100))^(1//2)`
`=10(1+0.01)^(1//2)`
`~= 10(1+(0.01)/(2))`
`=10(1+0.005)`
`=10.05`
Discussion
53.

Find the equation of trajectory of a particle whose velocity components are `v_(x)=2x+1, v_(y)=2y+3` Given that particle starts from rest from origin.

Answer» `v_(x)=2x+1`,
`(dx)/(dt)=2x+1`
`int_(0)^(x) (dx)/(2x+1)=int_(0)^(t)dt`
`(1)/(2)ln(2x+1)=t " "` ....(1)
`v_(y)=2y+3`
`int _(0)^(y)(dy)/(2y+3)=int_(0)^(t)dt`
`(1)/(2) "ln "((2y+3))/(3)=t" "` ...(2)
from (1) and (2)
`(1)/(2) ln (2x+1)=(1)/(2) "ln" ((2y+3))/(3)`
`2x+1=((2y+3))/(3)`
`y=3x`
Discussion
54.

If `vec(r)=[ucos theta(hat(i))+u sin theta(hat(j))]t+(1)/(2)g(-hat(j))t^(2)` then calculate equation of trajectory.

Answer» `x=u cos theta t` & `y=u sin theta t-(1)/(2)g t^(2)`
`y=usintheta ((x)/(u)cos theta)-(1)/(2)g((x)/(u cos theta))^(2)`
`y=x tan theta-(gx^(2))/(2u^(2)cos^(2) theta)`
Discussion
55.

A body is moving vertically upwards under gravity such that its position from ground is given as `y=ut-(1)/(2)g t^(2)` . Find the max height reached by body.

Answer» `(dy)/(dt)=u-g t=0 ,t=(u)/(g)`
`(d^(2)y)/(dx^(2))=-glt0` (which is negative). So we get maximum height`y_(max)at" " t=(u)/(g)sec`.
`y_(max)=u((u)/(g))-(1)/(2)g((u)/(g))^(2)`
`(u^(2))/(2g)`
Discussion
56.

Find a vector `vec(F)` of magnitude 50N parallel to `-4hat(i)+3hat(j)`.

Answer» `vec(F)=50xx((-40hat(i)+3hat(j)))/(5)=-40hat(i)+30hat(j)`
Discussion
57.

A particle is moving with speed 6m/s along the direction of `2hat(i)+2hat(j)-hat(k)` find the velocity vector of a particle ?

Answer» `vec(v)=|vec(v)|hat(v)=6((2hat(i)+2hat(j)-hat(k)))/(3)=2(2hat(i)+2hat(j)-hat(k))`
Discussion
58.

Find the derivative of `y(x)=x^(3)//(x+1)^(2)` with respect to x .

Answer» We can rewrite this function as `y(x)=x^(3)(x+1)^(-2)`and apply Equation()
`(dy)/(dx)=(x+1)^(-2) (d)/(dx)(x^(3))+x^(3)(d)/(dx)(x+1)^(-2)`
`=(x+1)^(-2)3x^(2)+x^(3)(-2)(x+1)^(-3)`
`(dy)/(dx)=(3x^(2))/((x+1)^(2))-(2x^(3))/((x+1)^(3))`
Discussion