Physics, asked by anantrajusharma, 9 months ago

please solve it I have to show the solutions upto 2:00 pm
class 11 physics​

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Answers

Answered by Kshitij2299
6

Refer the attachment

may you get the HW done on time : )

Hope it helps

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Answered by shadowsabers03
4

1. Let \sf{t_1} and \sf{t_2} be the time taken for travelling distances \sf{x_1} and \sf{x_2} with velocities \sf{v_1} and \sf{v_2} respectively. Hence we have,

  • \sf{t_1=\dfrac{x_1}{v_1}}

  • \sf{t_2=\dfrac{x_2}{v_2}}

Hence average velocity is,

\longrightarrow\sf{Average\ Velocity=\dfrac{Total\ Displacement}{Total\ Time}}

\longrightarrow\sf{\bar v=\dfrac{x_1+x_2}{t_1+t_2}}

\longrightarrow\sf{\bar v=\dfrac{x_1+x_2}{\left(\dfrac{x_1}{v_1}+\dfrac{x_2}{v_2}\right)}}

\longrightarrow\sf{\underline{\underline{\bar v=\dfrac{v_1v_2(x_1+x_2)}{x_1v_2+x_2v_1}}}}

2. The distance travelled after one full revolution is equal to the perimeter of the circular path of diameter d = 14 cm.

Hence distance travelled is,

\longrightarrow\sf{D=\pi d}

Taking \sf{\pi=\dfrac{22}{7},}

\longrightarrow\sf{D=\dfrac{22}{7}\times14}

\longrightarrow\sf{\underline{\underline{D=44\ m}}}

Since the particle reaches initial position after one full revolution, its displacement is 0 m.

\longrightarrow\sf{\underline{\underline{s=0\ m}}}

3. We know that velocity is first derivative of displacement with respect to time.

\longrightarrow\sf{v=\dfrac{dx}{dt}}

\longrightarrow\sf{v=\dfrac{d}{dt}\,\left[2-5t+6t^2\right]}

\longrightarrow\sf{v=12t-5}

Since initial velocity is the velocity of particle at t = 0,

\longrightarrow\sf{v(0)=12(0)-5}

\longrightarrow\underline{\underline{\sf{v(0)=-5\ m\,s^{-1}}}}

4. At t = 0,

\longrightarrow\sf{x(0)=4(0)^2-15(0)+20}

\longrightarrow\underline{\underline{\sf{x(0)=20\ m}}}

The velocity of the particle is,

\longrightarrow\sf{v=\dfrac{dx}{dt}}

\longrightarrow\sf{v=\dfrac{d}{dt}\,\left[4t^2-15t+20\right]}

\longrightarrow\sf{v=8t-15}

At t = 0,

\longrightarrow\sf{v(0)=8(0)-15}

\longrightarrow\underline{\underline{\sf{v(0)=-15\ m\,s^{-1}}}}

The acceleration of the particle is,

\longrightarrow\sf{a=\dfrac{dv}{dt}}

\longrightarrow\sf{a=\dfrac{d}{dt}\,[8t-15]}

\longrightarrow\sf{a=8\ m\,s^{-2}}

At t = 0,

\longrightarrow\underline{\underline{\sf{a=8\ m\,s^{-2}}}}

5. Let x (in km) be the distance between A and B.

Time taken to travel from A to B, \sf{t_1=\dfrac{x}{40}\ h.}

Time taken to return back, \sf{t_2=\dfrac{x}{60}\ h.}

Hence average speed is,

\longrightarrow\sf{v=\dfrac{x+x}{t_1+t_2}}

\longrightarrow\sf{v=\dfrac{2x}{\left(\dfrac{x}{40}+\dfrac{x}{60}\right)}}

\longrightarrow\sf{v=\dfrac{2x\cdot2400}{100x}}

\longrightarrow\underline{\underline{\sf{v=48\ km\,h^{-1}}}}

Since the car reaches back to its initial position, it has zero displacement. Hence the average velocity is 0 km/h.

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