Question

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

Answer 1

Refer the attachment

may you get the HW done on time : )

Hope it helps

Answer 2

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.