6. The displacement x of a particle varies with time t as x = ae^−αt + be^βt, where a, b, α and β are positive constants. The velocity of the particle will

1. 1. be independent of α and β

   
   
   

   2. drop to zero when α = β

   
   
   

   3. go on decreasing with time

   
   
   

   4. go on increasing with time

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given x = ae^–αt + be^βt
   Velocity, v = dx/dt = –aαe^–αt + bβe^βt
   

   =	aα	= bβeβt
   	eαt

   
   i.e., go on increasing with time.

   Correct Option: D

   Given x = ae^–αt + be^βt
   Velocity, v = dx/dt = –aαe^–αt + bβe^βt
   

   =	aα	= bβeβt
   	eαt

   
   i.e., go on increasing with time.

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7. 
   A particle shows distance - time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point:

1. 1. B

   
   
   

   2. C

   
   
   

   3. D

   
   
   

   4. A

   
   
   

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1. View Hint View Answer Discuss in Forum

   The slope of the graph ds/dt is maximum at C and hence the instantaneous velocity is maximum at C.

   Correct Option: B

   The slope of the graph ds/dt is maximum at C and hence the instantaneous velocity is maximum at C.

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8. If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of

1. 1. 2880 m

   
   
   

   2. 1440 m

   
   
   

   3. 400 m

   
   
   

   4. 20 m

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Initial velocity of car (u) = 0
   Final velocity of car (v) = 144 km/hr = 40 m/s
   Time taken = 20 s
   We know that, v = u + at
   40 = a × 20 ⇒ a = 2 m/s²
   Also, v² – u² = 2as
   

   ⇒ s =	v² – u²
   	2a

   
   

   ⇒ s =	(40)² – (0)²	=	1600	= 400 m
   	2 × 2		4

   Correct Option: C

   ​Initial velocity of car (u) = 0
   Final velocity of car (v) = 144 km/hr = 40 m/s
   Time taken = 20 s
   We know that, v = u + at
   40 = a × 20 ⇒ a = 2 m/s²
   Also, v² – u² = 2as
   

   ⇒ s =	v² – u²
   	2a

   
   

   ⇒ s =	(40)² – (0)²	=	1600	= 400 m
   	2 × 2		4

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9. Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be:

1. 1. t1t2

      t2 - t1

   
   
   

   2. t1t2

      t2 + t1

   
   
   

   3. t₁ - t₂

   
   
   

   4. t1 + t2

      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Velocity of preeti w.r.t. elevator v1 =	d
   	t1

   
   

   Velocity of elevator w.r.t. ground then velocity of preeti w.r.t. ground v2 =	d
   	t2

   
   v = v₁ + v₂
   

   d	=	d	+	d
   t		t1		t2

   
   

   1	=	1	+	1
   t		t1		t2

   
   

   ∴ t =	t1t2	(time taken by preeti to walk up on the moving escalator)
   	(t1 + t2)

   Correct Option: B

   Velocity of preeti w.r.t. elevator v1 =	d
   	t1

   
   

   Velocity of elevator w.r.t. ground then velocity of preeti w.r.t. ground v2 =	d
   	t2

   
   v = v₁ + v₂
   

   d	=	d	+	d
   t		t1		t2

   
   

   1	=	1	+	1
   t		t1		t2

   
   

   ∴ t =	t1t2	(time taken by preeti to walk up on the moving escalator)
   	(t1 + t2)

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10. A car moves from X to Y with a uniform speed v_u and returns to Y with a uniform speed v_d. The average speed for this round trip is

1. 1. √v_uv_d

   
   
   

   2. vdvu

      vd + vu

   
   
   

   3. vd + vu

      2

   
   
   

   4. 2vdvu

      vd + vu

   
   
   

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1. View Hint View Answer Discuss in Forum

   Average speed =	total distance travelled
   	total time taken

   
   Let s be the distance from X to Y.
   

   ∴ Average speed =	s + s	=	2s
   	t1 + t2		s/vu + s/vd

   Correct Option: D

   Average speed =	total distance travelled
   	total time taken

   
   Let s be the distance from X to Y.
   

   ∴ Average speed =	s + s	=	2s
   	t1 + t2		s/vu + s/vd

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