Explanation
Given,
Velocity $${{v}_{1}}$$ achieve in time $${{t}_{1}}$$,due to acceleration, $${{a}_{1}}=0.2\,m{{s}^{-2}}$$
$$ {{v}_{1}}={{u}_{1}}+a{{t}_{1}} $$
$$ \Rightarrow {{v}_{1}}=0+0.2{{t}_{1}} $$
Velocity reduce to zero in time $${{t}_{2}}$$, due to retardation, $${{a}_{2}}=-0.4\,m{{s}^{-2}}$$
$$ {{v}_{2}}={{v}_{1}}+a{{t}_{2}} $$
$$ \Rightarrow 0={{v}_{1}}-0.4{{t}_{2}} $$
Add time
$$ {{t}_{1}}+{{t}_{2}}=\dfrac{{{v}_{1}}}{0.2}+\dfrac{{{v}_{1}}}{0.4} $$
$$ \Rightarrow 30\times 60={{v}_{1}}\times 7.5 $$
$$ \Rightarrow {{v}_{1}}=240\,m/s $$
Apply kinematic equation
$$ {{v}^{2}}-{{u}^{2}}=2as $$
$$ s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a} $$
Total distance $${{S}_{1}}+{{S}_{2}}=\dfrac{{{v}_{1}}^{2}}{2{{a}_{1}}}+\dfrac{{{v}_{1}}^{2}}{2{{a}_{2}}}=\dfrac{{{240}^{2}}}{2\times 0.2}+\dfrac{{{240}^{2}}}{2\times 0.4}=216\,km$$
A ball dropped from the top of the tower falls first half height of the tower in 10s. The total time spent by a ball in the air is $$\left[ {Take\;g = 10m/{s^2}} \right]$$
Let the height of tower is h. Now considering the equation as
$$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$
$$ h=ut+\dfrac{1}{2}a{{t}^{2}}...........(1) $$
$$ here,\,\,h=\dfrac{h}{2} $$
In 10 sec, object is reaching half the height of tower.
Initially the body is rest so u = 0
It is also given that a = 10 ms-2
Time to reach first half height of tower t = 10 sec
Put the values in (1)
$$ \dfrac{h}{2}=0+\dfrac{1}{2}\times 10\times {{10}^{2}} $$
$$ h=1000\,\,m $$
The total time spent by a ball in the air is given by the relation as
$$ 1000=\dfrac{1}{2}\times 10\times {{t}^{2}} $$
$$ t=10\sqrt{2} $$
$$ t=14.14\,\sec $$
A body moving along the positive $$x-axis$$ with uniform acceleration of $$ - 4m{s^{ - 2}}$$. Its velocity at $$x=0$$ is $$ 10m{s^{- 1}}$$. The time taken by the body to reach a point at $$ x = 12cm$$ is:
Let,
Constant acceleration, $$a$$
Initial velocity $$u$$
Apply kinematic Equation
$$v=u+at$$
$$a=\dfrac{v-u}{t}$$
Compare acceleration at both event at time, $${{t}_{1}}\,\And \,{{t}_{2}}$$
$$\dfrac{{{v}_{2}}-u}{{{t}_{1}}}=\dfrac{{{v}_{3}}-u}{{{t}_{2}}}$$
$$\Rightarrow \dfrac{30-u}{2}=\dfrac{60-u}{4}$$
$$\Rightarrow u=0$$
Initial velocity $$u=zero$$
The acceleration when the mass is going up is given as,
$${v^2} = {u^2} + 2as$$
$${\left( {20} \right)^2} = {\left( 0 \right)^2} + 2as$$
$$a = \dfrac{{200}}{s}$$ (1)
The acceleration when the mass is going down is given as,
$$s = \dfrac{1}{2}a{t^2}$$
$$s = \dfrac{1}{2}a{\left( {10} \right)^2}$$
$$a = \dfrac{s}{{50}}$$ (2)
From equation (1) and (2), we get
$$\dfrac{{200}}{s} = \dfrac{s}{{50}}$$
$$s = 100\;{\rm{m}}$$
Substitute the value of $$s$$ in the equation (2), we get
$$a = \dfrac{{100}}{{50}}$$
$$ = 2\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$
The weight is given as,
$$W = 0.5 \times 2$$
$$ = 1\;{\rm{N}}$$
Thus, the weight of the body on that planet is $$1\;{\rm{N}}$$.
Force on crate, $$F=\,575\,N$$
Mass of crate, $$m=50\,kg$$
Coefficient of friction, $$\mu =0.25$$
From the equilibrium of forces
$$ma=F-\mu mg$$
$$a=\dfrac{F-\mu mg}{m}$$
$$a=\dfrac{575-0.25\times 50\times 9.81}{50}$$
$$a=9.047\,\,m/{{s}^{2}}$$
Acceleration of block is $$9.047\,\,m/{{s}^{2}}$$
Let distance covered by A be S1
Let distance covered by B be S2
B will catch A when S1 = S2
S = ut + 1/2 * a * t^2
Therefore,
40t + (1/2 * 0 * t^2) = 0t + 1/2 * 4 * t^2
40t = 2t^2
20t = t^2
Therefore, t = 20
Given that,
Mass $${{m}_{1}}=3\,kg$$
Mass $${{m}_{2}}=2\,kg$$
Now, the initial acceleration a of the system
$$ a=\dfrac{3g-2g}{3+2} $$
$$ a=\dfrac{g}{5} $$
At $$5\ sec$$
Now, from equation of motion
$$ v=u+at $$
$$ v=0+\dfrac{g}{5}\times 5 $$
$$ v=9.8\,m/s $$
When the string breaks at $$t= 5\ s$$,
The mass $$2\ kg$$ more under gravity
So, height is
$$ h=\dfrac{{{v}^{2}}}{2g} $$
$$ h=\dfrac{{{\left( 9.8 \right)}^{2}}}{2\times 9.8} $$
$$ h=4.9\,m $$
Hence, the height is $$4.9\ m$$
Initial velocity, $$u=\left( 2\hat{j}+\hat{k} \right)\,m/s$$
Force, $$F=\left( 2\hat{i}+\hat{j}+3\hat{k} \right)\,N$$
$$\vec{a}=\dfrac{{\vec{F}}}{m}=\dfrac{2\hat{i}+\hat{j}+3\hat{k}}{1}=\left( 2\hat{i}+\hat{j}+3\hat{k} \right)\,m/{{s}^{2}}$$
$$ \vec{v}=\vec{u}+\vec{a}t $$
$$ \vec{v}=\left( 2\hat{j}+\hat{k} \right)+\left( 2\hat{i}+\hat{j}+3\hat{k} \right)\times 2 $$
$$ \vec{v}=\left( 4\hat{i}+4\hat{j}+7\hat{k} \right)\,m/s $$
$$ \left| {\vec{v}} \right|=\sqrt{{{4}^{2}}+{{4}^{2}}+{{7}^{2}}}=9\,m/s $$
Final velocity is $$9\,m/s$$
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