Maths M.C.Q (1 Marks)

At $t = 0{\left( {\frac{{ds}}{{dt}}} \right)_{t.0}} = - 3 = {v_1}$

Now, at $t = 1$, we get ${\left( {\frac{{ds}}{{dt}}} \right)_{t = 1}} = 4 - 3 = 1 = {v_2}$

Hence rate of velocity $ = {v_2} - {v_1} = 1 - ( - 3) = 4$

Aliter : Given that $s = 2{t^2} - 3t + 1$

$\frac{{ds}}{{dt}} = 4t - 3 $ ( velocity), Again $\frac{{{d^2}s}}{{d{t^2}}} = 4$ (acceleration).

and $\frac{{ds}}{{dt}} = $ velocity $2t - 2 $ .....$(ii)$

Put $s = 15$ in $(i),$ we get ${t^2} - 2t - 15 = 0$

or $(t - 5)(t + 3) = 0$

$\therefore t = 5$

Hence velocity of car $ = v = 2(5) - 2 = 8\,\,km/h$.

Therefore, $S = \frac{{{u^2}}}{{2g}} = 12250.$

(Increasing the rate/sec is called the velocity)

$\frac{{da}}{{dt}} = 5$ .....$(i)$

Where $ a$ is distance and  $t$ is time.

But if $a$ is edge of a cube, then $V = {a^3}$

Differentiating w.r.t. time $t,$ so

$\frac{{dV}}{{dt}} = 3{a^2}\frac{{da}}{{dt}} = 3{a^2}.5\, = \,15{a^2} = 15 \times {(12)^2}$

$ = 2160\,\,c{m^3}/\sec $ ( $\because$ edge $a = 12\,cm)$.

==> $2\frac{{{d^2}s}}{{d{t^2}}} = \frac{{dv}}{{dt}} + t.\frac{{{d^2}v}}{{d{t^2}}} + \frac{{dv}}{{dt}}$

But $\frac{{dv}}{{dt}} =$ acceleration $(a)$

==> $2a = a + t.\frac{{da}}{{dt}} + a$ ==> $\frac{{da}}{{dt}} = 0$ or $xy = 4 \times 4 = 16$

But for whole notation $t = 0$ is impossible so that $\frac{{da}}{{dt}} = 0$ $i.e.,$ $a$ is constant.

Therefore, velocity $\frac{{ds}}{{dt}} = 15 - 4t$

==> ${\left( {\frac{{ds}}{{dt}}} \right)_{t = 0}} = 15$ and ${\left( {\frac{{ds}}{{dt}}} \right)_{t = 3}} = 3$

Therefore, average $ = \frac{{15 + 3}}{2} = 9$.

But $v = 0$ (at maximum height),    $\therefore t = \frac{{10}}{6}$

Therefore, the stone will return in $2 \times \frac{{10}}{6} = \frac{{10}}{3}\sec .$

The time taken by second stone to secure maximum height is, $t = \frac{{9.8}}{{9.8}} =  1\,\,sec. $

Therefore, in $2\, sec$, second stone will come back to the ground. Hence height $s = a{t^2} + bt + 6$.

==> $\frac{{ds}}{{dt}} = u - 9.8t = v$

When stone reached the maximum height, then $v = 0$

==> $u - 9.8t = 0 \Rightarrow u = 9.8\,t$

But time $t = 5\,sec$

So the value of $u = 9.8 \times 5 = 49.0\,m/sec$

Hence initial velocity $ = 49\,m/sec.$

Again $\frac{{{d^2}s}}{{d{t^2}}} = 12t - 18  =$ acceleration

If acceleration becomes zero, then $0 = 12t - 18$ ==> $t = \frac{3}{2}\sec .$

Hence acceleration will be zero after $\frac{3}{2}\,sec.$

$=$ Distance travelled in $3\, sec -$ Distance travelled in $2\, sec.$

$ = (9 + 24 + 12) - (4 + 16 + 12) = 45 - 32 = 13$.

$ \Rightarrow 2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} = 0$ .....$(ii)$

$x = 8$

Therefore by $(i)$ and $(ii),$

$\frac{{dy}}{{dt}} = - \frac{{16}}{6} = - \frac{8}{3}cm/\sec .$

$ = \frac{8}{3}cm/\sec $.

$\therefore \frac{{dA}}{{dt}} = 2\pi R\frac{{dR}}{{dt}} = 1.2\pi \,c{m^2}$.

Differentiating w.r.t. time, we get

Velocity $(v)= 2at + b$ ……$(ii)$

After $4\,sec,$ $v = 0$ and distance $s = 16\,metres$

$\therefore 0 = 2a \times 4 + b \Rightarrow 8a + b = 0$ …..$(iii)$

and $16 = 16a + 4b + 6$==> $16 = 16a + 4( - 8a) + 6$

$\therefore $ $a = - \frac{5}{8}$

But retardation in its motion is, $2a = \frac{{ - 5}}{4}m/{\sec ^2}$

$\therefore $ Retardation $ = \frac{5}{4}m/{s^2}$ (Retardation itself means $-ve$).

When particle will come to rest, then $v = 0$

==> $3{t^2} - 22t - 45 = 0 \Rightarrow t = 9,\,\,\left( {{\rm{since}}\,\,{\rm{t}} \ne - \frac{5}{3}} \right)$.

Here terminal velocity $v = 0$ and $t = 3\,sec$

$u = 9.8(3) = 29.4 \,m/sec$.

==> ${\left( {\frac{{dS}}{{dt}}} \right)_{r = 10}} = 2.\pi \times 10 \times 3 = 60\,\pi \,c{m^2}/sec.$

Its velocity $v = \frac{{ds}}{{dt}} = 13.8 - 9.8\,t$

Hence ${\left( {\frac{{ds}}{{dt}}} \right)_{t = 1}} = 13.8 - 9.8 \times 1 = 4.0 = 4\,m/\sec .$

Now velocity $v = - 8t + 2$ and its acceleration $a = - 8$

So ${\left( {\frac{{ds}}{{dt}}} \right)_{t = 1/2}} = - 8 \times \frac{1}{2} + 2 = - 2$ and

${\left( {\frac{{{d^2}s}}{{d{t^2}}}} \right)_{t = 1/2}} = - 8$.

From figure, $\frac{x}{2} = \frac{{x + y}}{6} \Rightarrow 4x = 2y \Rightarrow x = \frac{1}{2}y$

Hence $\frac{{dx}}{{dt}} = \frac{1}{2}\frac{{dy}}{{dt}} = \frac{5}{2}metre/hour$.

Differentiate $(i)$ $w.r.t.$ $t,$ we get

$2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} = 0$ …..$(ii)$

Here $x = 4$ and $\frac{{dx}}{{dt}} = 1.5$

From $(i),$ ${4^2} + {y^2} = 25 \Rightarrow y = 3$

$\therefore $ From $(ii),$ $2(4)(1.5) + 2(3)\frac{{dy}}{{dt}} = 0$

So, $\frac{{dy}}{{dt}} = - 2\,m/\sec $

Hence, length of the highest point decreases at the rate of $2 \,m/sec.$

Area of circle $ = \pi {r^2}$, $A = \pi {r^2}$

==> $\frac{{dA}}{{dt}} = 2\pi r.\frac{{dr}}{{dt}}$==> $\frac{{dA}}{{dt}} = 2\pi \times 10 \times 3.5$

==> $\frac{{dA}}{{dt}} = 220c{m^2}/sec$.

$V = {a^3}$.

$\therefore \frac{{dV}}{{dt}} = 3{a^2}\frac{{da}}{{dt}} = 3{a^2} \times 60 = 180{a^2} = 180 \times {(90)^2}$

$ = 1458000 \, c{m^3}/sec$.

$\therefore $ $\frac{{ds}}{{dt}} = a\cos t - 2b\sin 2t$

$\frac{{{d^2}s}}{{d{t^2}}} = - a\sin t - 4b\cos 2t$

At $t = 0$, $\frac{{{d^2}s}}{{d{t^2}}} = - a\sin 0^\circ - 4b\cos 0^\circ = - 4b$.

$v = \frac{{dS}}{{dt}} = 0 + 48 - 3{t^2}$

when direction of motion reverses, $v = 0$

$48 - 3{t^2} = 0$ $ \Rightarrow $ $t = - 4,\;4$;

${(S)_4} = 6 + 192 - 64$  $i.e.,$ $S = 134$.

Differentiate with respect to $t,$

$\frac{{dV}}{{dt}} = \frac{4}{3}\pi 3{r^2}.\frac{{dr}}{{dt}}$ ==> $\frac{{dr}}{{dt}} = \frac{1}{{4\pi {r^2}}}.\frac{{dV}}{{dt}}$

$\frac{{dr}}{{dt}} = \frac{1}{{4 \times \pi \times 15 \times 15}} \times 900$

$ \Rightarrow $ $\frac{{dr}}{{dt}} = \frac{1}{\pi } = \frac{7}{{22}}$.

${a_x}$ is acceleration in $x-$ axis

$\frac{{{d^2}y}}{{d{t^2}}} = - b{a^2}\sin at \Rightarrow {a_y} = - {a^2}y$

Hence, ${a_y}$ changes as  $ y  $ changes.

$\frac{{dP}}{{dt}} = 2\pi .\,\frac{{dr}}{{dt}}$ and $A = \pi {r^2}$

$\frac{{dA}}{{dt}} = 2\pi r.\,\,\frac{{dr}}{{dt}}$

==> $\frac{{dA}}{{dt}} = \frac{{dP}}{{dt}} \times r$==> $\frac{{dA}}{{dt}} \propto r$.

${(v)_{t = 2}} = 80 - 64 = 16\,m/sec$.

==> $\frac{{ds}}{{dt}} = 1 + 12t - 3{t^2}$ ==> $v = 5\,cm/\sec $

==> $\frac{{{d^2}s}}{{d{t^2}}} = 0$ ==> $12 - 6t = 0 \Rightarrow t = 2$.

The body will be stopped when $v = 0$

==> $6t - 8 = 0$ ==> $t = 4/3$.

==> $\frac{{dy}}{{dx}} = \frac{1}{2}{({x^2} + 16)^{ - 1/2}}(2x)$ and

$\frac{{dz}}{{dx}} = \frac{{x - 1 - x}}{{{{(x - 1)}^2}}} = \frac{{ - 1}}{{{{(x - 1)}^2}}}$

$\therefore$  $\frac{{dy}}{{dz}} = \frac{{ - x}}{{\sqrt {{x^2} + 16} }}\,\frac{1}{{1/{{(x - 1)}^2}}}$

${\left( {\frac{{dy}}{{dz}}} \right)_{x = 3}} = \frac{{ - 3{{(2)}^2}}}{5} = \frac{{ - 12}}{5}$.

==> $0 + b\left( {2v.\,\frac{{dv}}{{dt}}} \right) = 2x\,.\,\frac{{dx}}{{dt}}$

==> $v.b\frac{{dv}}{{dt}} = x\,.\,\frac{{dx}}{{dt}}$

==> $\frac{{dv}}{{dt}} = \frac{x}{b}$ ,    $\left(\because {\frac{{dx}}{{dt}} = v} \right)$

$A = \frac{{\sqrt 3 }}{4}{x^2} \Rightarrow \frac{{dA}}{{dt}} = \frac{{\sqrt 3 }}{4}2x\frac{{dx}}{{dt}}$

Here, $x = 10\,\,cm$ and $\frac{{dx}}{{dt}} = 2\,\,cm/sec$

==> $A = 10\sqrt 3 \,\,sq.\, unit/sec$

$\therefore$  $\frac{{ds}}{{dt}} = 4\pi \times 2r\frac{{dr}}{{dt}}  = 8\pi r \times 2 = 16\pi r$

==> $\frac{{ds}}{{dt}} \propto r$.

==> ${t^2} - 6t - 7 = 0$, $t = 7\,sec.$

At $t = 7\sec $., ${v_1} = \frac{d}{{dt}}(10 + 6t) = 6\,cm/sec$

At $t = 7\sec .$ ${v_2} = \frac{d}{{dt}}(3 + {t^2}) = 2t = 2 \times 7 = 14\,cm/sec$

$\therefore$ Resultant velocity $= {v_2} - {v_1} = 14 - 6 = 8\,cm/sec$

acceleration in direction of $ y-axis $$ = \frac{{{d^2}y}}{{d{t^2}}} = 2b$

$\therefore$ Resultant acceleration is $=\sqrt {{{( - 2c)}^2} + {{(2b)}^2}} = 2\sqrt {{b^2} + {c^2}} $

$\therefore$  $v = \frac{{ds}}{{dt}} = 3{t^2} - 6t$, $a = \frac{{{d^2}s}}{{d{t^2}}} = 6t - 6$

Acceleration is zero, if $6t - 6 = 0$==> $t = 1$

$\therefore$  Required velocity of particle at $t = 1$ is $v = 3{(1)^2} - 6(1)$

$ \Rightarrow \,v = - 3$.

Also $a = \frac{{dv}}{{dt}} = \frac{{ - 1}}{{2 \times 2{{(t)}^{3/2}}}}$

==> $a \propto \frac{1}{{t\sqrt t }}$ or $a \propto {v^3}$.

$35\, cc/  min$ $= 4\pi {(7)^2}\frac{{dr}}{{dt}} \Rightarrow \frac{{dr}}{{dt}} = \frac{5}{{28\pi }}$

Surface area, $S =  4\pi {r^2}$ 

$\frac{{dS}}{{dt}} = 8\pi r\frac{{dr}}{{dt}} = \frac{{8\pi .7.5}}{{28\pi }} = 10\,$ $cm^2/min$.

Hence $ a $ is constant.

$\frac{{\partial V}}{{\partial t}} = \pi \,[2(4)\,(3)\,(6)\, + \,{(4)^2}( - 4)] $

$= \pi \,[144 - 64] = 80\,\pi {\rm{ }}cu\,m/sec$.

==> $2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} = 0$ ==> $x.3 + y.( - 4) = 0$

==> $x = \frac{4}{3}y$.

Thus, ${\left( {\frac{4}{3}y} \right)^2} + {y^2} = {10^2} \Rightarrow y = 6\,m.$

==> $\frac{{dV}}{{dt}} = 4\pi {r^2}\frac{{dr}}{{dt}}$

==> $f''(1) = 3 + 2{\log _e}1$

$\therefore \frac{{dS}}{{dt}} = 8\pi r\frac{{dr}}{{dt}} = 8\pi \times 8 \times \frac{5}{{32\pi }} = 10$.

As stone is thrown upwards, so acceleration $ = 1.6$.