physics ncert solutions class 11th

This is a multiple choice answer as classified in NCERT Exemplar

(c) Time taken to travel first half distance t₁= $\frac{l/2}{v_1}=\frac{l}{2v_1}$

Time taken to travel second half distance t₂= $\frac{l}{2v_2}$

Total time =t₁+t₂

formula for average velocity will be =  $\frac{l}{2v_1}+\frac{l}{2v_2}=\frac{l}{2}\left. [\frac{1}{v_1}+\frac{1}{v_2}\right]$

formula for average velocity will be= v_av=average speed= total distance/total time = $\frac{2v_1+v_2}{v_{1+}{v}_2}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) For maximum and minimum displacement we have to keep in mind the magnitude and direction of maximum velocity.

As maximum velocity in positive direction is v_o maximum velocity in opposite direction is also v_o.

Maximum displacement in one direction = v_oT

Maximum displacement in opposite direction=-v_oT

Hence -v_oToT

This is a multiple choice answer as classified in NCERT Exemplar

(a) As the lift when coming downward directions displacement will be negative. We have to see whether the motion is accelerating or retarding.

We know that due to downward motion displacement will be negative, when the lift reaches 4^th floor is about to stop hence motion is retarding in nature.

As displacement is in negative direction . velocity will also be negative i.e v<0  

This is a multiple choice answer as classified in NCERT Exemplar

(b) To making it vanish one part must cancel other part which is only possible in graph b.

As there are opposite velocities in the interval 0 to T hence average velocity can vanish in b .

This is a short answer type question as classified in NCERT Exemplar

let initial velocity V_i

Distance travelled in time t= x_i

For the graph

tan $\theta$ = v_i/x_i= v_i-v/x

V= -v_ix/x_i+V_o

Acceleration, a = dv/dx= -v_idx/x_idt

So a= -v_iv/x_i= $\frac{{-v}_i}{x_i}(\frac{{-v}_i}{x_i}x+v_i)$

This is a short answer type question as classified in NCERT Exemplar

When the ball dropped from the building u₁=0, u₂=40m/s

Velocity of the dropped ball after time t

V₁=u₁+gt

V₁ = gt

For ball thrown up u₂=40m/s

Velocity of the ball after time t

V₂=u₂-gt

=40-gt

Relative velocity =v₁-v₂

=gt- [-40-gt]=40m/s

This is a short answer type question as classified in NCERT Exemplar

When the ball dropped from the building u₁=0, u₂=40m/s

Velocity of the dropped ball after time t

V₁=u₁+gt

V₁ = gt

For ball thrown up u₂=40m/s

Velocity of the ball after time t

V₂=u₂-gt

=40-gt

Relative velocity =v₁-v₂

=gt- [-40-gt]=40m/s

This is a short answer type question as classified in NCERT Exemplar

speed of first car = 18km/h

Speed of second car = 27km/hr relative with respect to each other is 18+27=45km/h

Distance between cars = 36km

Time = $\frac{distancebetweencars}{relativespeed}=$ 36/45=0.8h

Distance covered by bird = 36 (0.8)=28.8km

This is a short answer type question as classified in NCERT Exemplar

(a) x (t)=x₀ (1- $e^{-\gamma t}$ )

V (t)=dx/dt= x (t)=x_{0 $\gamma$} ( $e^{-\gamma t}$ )

A (t)= x ( $\infty ,$ t)=x₀ ( ${\gamma }^2{e}^{-\gamma t}$ )

(b) when t=o x (t)= x (t)=x₀ (1- $e^{-0}$ )= x (t)=x₀ (1-1)=0

x (t) is maximum when t= $\infty$

x (t) is maximum when t= $0$

v (t) is maximum when t= $0$ , v (0)=x_{0 $\gamma$}

v (t) is maximum when t= $\infty ,$  v ( $\infty$ )=0

a (t) is maximum when t= $\infty$ , a ( $\infty$ )=0

a (t) is maximum when t= $0$ , a ( $0$ )=-x_{0 $\gamma$} ²

This is a short answer type question as classified in NCERT Exemplar

It is clear from the graph that displacement x is positive throughout . ball is dropped from height and its velocity increases in downward direction due to gravity pull. In this condition v isnegative but acceleration of the ball equal to acceleration due to gravity.

velocity time graph

Acceleration time graph