Physics NCERT Exemplar Solutions Class 11th Chapter Three

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(a, c) as we know restoring force is F=-kx

Potential energy of the spring = 1/2kx²

The restoring force is central hence, when particle released it will execute SHM about equilibrium position

Also acceleration= f/m=-kx/m

But at equilibrium position when x=o, a also 0 and at equilibrium position potential energy converted into kinetic energy so speed will also maximum that time.

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(a, d) x= t-sint

Velocity v = dx/dt=

d/dt (t-sint)=

=1-cost

When cost =1, velocity v=0

V_max=1-cost_min = 1- (-1)=2

V_{min =}1- (cost)_max= 1-1=0

Hence v lies between 0 and 2

Acceleration a=dv/dt=-sint

When v=0 then cost =1

V_max= 1- (-1)=2, V_min=1-1=0

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(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.

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

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(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

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(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 = $\frac{2v_1+v_2}{v_{1+}{v}_2}$

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(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

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(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