Motion in A Straight Line Test - 54

Given that,

$$ a=b{{t}^{n}} $$

$$ \dfrac{dv}{dt}=b{{t}^{n}} $$

$$ dv=b{{t}^{n}}dt $$

$$ v=b\dfrac{{{t}^{n+1}}}{n+1}........(1) $$

$$ \dfrac{dr}{dt}=b\dfrac{{{t}^{n+1}}}{n+1} $$

$$ dr=\dfrac{b}{n+1}\int{{{t}^{n+1}}dt} $$

$$ r=\dfrac{b}{n+1}\dfrac{{{t}^{n+2}}}{n+2} $$

$$ r=\dfrac{Vt}{n+2}\,\,(from\,1) $$

$$ at\,\,t=\,1\,s $$

$$ V=(n+2)r $$

Given,

Mass of ship, $$M=3\times {{10}^{7}}\,kg$$

Force, $$F=5\times {{10}^{4}}\,N$$

Distance, $$s=3\,m$$

Acceleration, $$a=\dfrac{F}{M}=\dfrac{5\times {{10}^{4}}}{3\times {{10}^{7}}}=\dfrac{1}{600\,}\,m{{s}^{-2}}$$

Apply kinematic equation of motion

$$ {{v}^{2}}-{{u}^{2}}=2as $$

$$ v=\sqrt{2as}=\sqrt{2\times \dfrac{1}{600}\times 3}=0.1\,m{{s}^{-1}} $$

Hence, velocity of ship $$0.1\,m{{s}^{-1}}$$

 

$$u=1\dfrac { m }{ s } $$

It is given that, the body reaches the maximum height in 6s. Let u be the initial velocity. At maximum height the velocity is zero. Using the equation of motion as

$$ v=u+at $$

$$ 0=u-(10)(6) $$

$$ u=60\,m/s $$

Distance travelled in n^th second is given by:

$${{S}_{n}}=u+\left( \dfrac{a}{2} \right)\left( 2n-1 \right)$$

Displacement in 1^st second is $${{S}_{1}}=60-\left( \dfrac{10}{2} \right)(2-1)=55\,m$$

Displacement in 7^th second is $${{S}_{2}}=60-\left( \dfrac{10}{2} \right)(14-1)=-5\,m$$

So, $$\dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{55}{5}=\dfrac{11}{1}$$

The ratio is $$11:1$$.

The formula for time of flight is

$$T=\dfrac{2u\sin \theta }{g}$$

It is given that time is 2 seconds. So,

$$ 2=\dfrac{2u\sin \theta }{g} $$

$$ g=u\sin \theta  $$

Squaring both sides

$${{g}^{2}}={{u}^{2}}{{\sin }^{2}}\theta ........(1)$$

Height travelled is

$$ H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g} $$

$$ H=\dfrac{{{g}^{2}}}{2g}\,\,\,\,(from\,\,1) $$

$$ H=\dfrac{g}{2}=5\,m $$

Given that,

Distance $$s=45\,m$$

Initial velocity $$u=0$$

We know that,

By using equation of motion

  $$ {{v}^{2}}={{u}^{2}}+2as $$

 $$ {{v}^{2}}=0+2\times 9.8\times 45 $$

 $$ {{v}^{2}}=882\,m/s $$

 $$ v=29.7\,m/s $$

Hence, the minimum speed is $$29.7\ m/s$$

Distance covered by mat $$S = vt$$
                                         $$S_1 = 7t$$   ....(1)

Distance covered by bus $$S = \dfrac{1}{2} \,at^2$$
                                        $$S = \dfrac{1}{2} \times \dfrac{1}{2} \times t^2$$
                                        $$S_2 = \dfrac{1}{4} t^2$$

$$S_2 + 40 = S_1$$
$$\dfrac{1}{4} t^2 + 40 = 7t$$
$$t^2 + 160 = 28t$$
$$t^2 - 20t - 8t + 60 = 0$$
$$t(t - 20) - 8(t - 20) = 0$$

Relative disp $$= 40$$
$$S = ut + \dfrac{1}{2} \,at^2$$
$$40 = 7t + \dfrac{1}{2} \left(\dfrac{1}{2}\right) t^2$$
$$t^2 - 28t + 160 = 0$$
$$t^2 - 20t - 8t + 160 = 0$$
$$(t - 20)(t - 8)$$
$$t = 20$$ sec.