Physics Motion in Plane

The radius of circle, the period of revolution $T$ , initial position and sense of revolution are indicated in the figure. x-projection of the radius vector of rotating particle $P$ is

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Contributor-Level 10

ω = \frac{2 π}{T} = \frac{π}{5} r a d / s

\bar{θ} (a t t = 0) = 0^{0}

x = 5 \sin ? ω t

x = 5 \sin ? (\frac{π}{5} t)

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The velocity of boat m still water is 10 ms $m s^{- 1}$ . The river flow velocity is 5 ms $m s^{- 1} .$

If $_{}$ $v_{b}$ ,= velocity of boat with respect to river velocity, $v_{b} =$ $_{}$ actual velocity of boat (i.e. with respect to bank), $v_{r} =$ $_{}$ velocity of stream, then the value of $v_{b}$ $_{}$ is

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Contributor-Level 9

$v_{b} = v_{b r} + v,$

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If $a \cdot b = | a * b |$ , then the $θ$ angle between a and b will be

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Contributor-Level 9

$a \cdot b = | a * b | a b$ $cos θ = a b sin θ$

$\frac{\underline{a b}}{\underline{a b}} = \frac{s i n θ}{c o s θ}$

$tan θ = 1$

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A man who can swim with a speed of $2 k h / h$ wants to cross a river $200 m$ wide. If the river is flowing with a speed of $3 k m / h$ then in what minimum time can the man cross the river?

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Contributor-Level 10

Time (minimum) taken to cross the river

$t = \frac{W}{V_{s / r}} = \frac{200 * 10^{- 3}}{2}$

$100 * 10^{- 3} h$

$= 0.1 h = 6 m i n u t e s$

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Two non-zero vectors $\vec{A}$ and $\vec{B}$ are essentially perpendicular if

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For orthogonal vectors A.B = 0

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A particle is moving in a straight line with velocity (v) = (t – 4) m/s. What is the average speed between t = 0 to t = 8 second?

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Contributor-Level 10

v = t – 4

s =

s = 8 + 8 = 16m/s

V_avg= 16/8 = 8m/s

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The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by:

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Contributor-Level 10

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α * \frac{α}{2 β} - β * {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ - $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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A fighter jet is flying horizontally at a certain with a speed of 200ms^-1. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle (θ)with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of (θ)will be _____________.

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Contributor-Level 9

B → fighter jet

A → anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos q = 200

&nbs

...more

B → fighter jet

A → anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos q = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

less

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2 months ago

A person can throw a ball upto a maximum range of 100m. How high above the ground he can the same ball?

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Vishal Baghel

Contributor-Level 10

According to question, we can write

$R = \frac{u^{2} s i n 2 θ}{g} \Rightarrow u^{2} = g R_{m a x}, w h e r e θ = 45 °$

$\Rightarrow H_{m a x} = \frac{u^{2}}{2 g} = \frac{g R_{m a x}}{2 g} = \frac{R_{m a x}}{2} = \frac{100}{2} = 50 m$

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Motion of a particle in x – y plane is described by a set of following equations x = 4sin $(\frac{π}{2} - ω t)$ m and y = 4sin $(ω t) m .$ The path of a particle will be:

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