Motion in a Straight Line

This is a multiple choice answer as classified in NCERT Exemplar

(a, c, d) at point A graph will parallel to time axis hence, v=dx/dt=0. As the starting point is A hence, we can say that the particle is starting from rest.

At C, the graph changes slope, hence velocity also changes. As graph at C is almost parallel to time axis hence, we can say that velocity vanishes.

Velocity is 0. So particle at rest that time. As slope of D will be greater than E . so velocity will be greater at D.

This is a multiple choice answer as classified in NCERT Exemplar

(c) in this situation we have to find net velocity with respect to the earth that will be equal to velocity of the girl plus velocity of escalator.

Let displacement is L the

Velocity of  girl v_g= L/t₁

Velocity of escalator v_e= L/t₂

net velocity = velocity of escalator +velocity of girl

V_g=l/t₁  and V_e=l/t₂

So net velocity = L/t₁+ L/t₂

So t= $\frac{t_1{t}_2}{t_1+t_2}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) x = (t-2)²

V= dx/dt= 2 (t-2)

And acceleration a = 2

When t =0 v= -4m/s

When t= 2s v= 0m/s

When t= 4s v= 4m/s

Distance travelled = area of graph =area of OAC+area of ABD= $\frac{4*2}{2}+\frac{1}{2}*2*4$ =8m

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