1. A car moving with a speed of 40 km/h can be stopped by applying brakes at least after 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance?

1. 1. 8 m

   
   
   

   2. 6 m

   
   
   

   3. 4 m

   
   
   

   4. 2 m

   
   
   

---

1. View Hint View Answer Discuss in Forum

   v² – u² = 2as
   

   ⇒ a =	v² - u²
   	2s

   
   

   =	- u1²	, where u1 = 40 km/h
   	2s

   
   For same retarding force s ∝ u²
   

   ∵	s2	=	u2²	⇒	s2	=		80		²	= 4
   	s1		u1²		s1			40

   
   ∴ s₂ = 4s₁ = 8m
   If  F is retarding force and s the stopping distance, then 1/2 mv² = Fs
   For same retarding force, s α v²
   

   ∴

   s2

   =

   v2

   ²

   =

   80 km/h

   ²

   = 4

   s1

   v1

   40 km/h

   
   ∴ s₂ = 4s₁ = 4 × 2 = 8m

   Correct Option: A

   v² – u² = 2as
   

   ⇒ a =	v² - u²
   	2s

   
   

   =	- u1²	, where u1 = 40 km/h
   	2s

   
   For same retarding force s ∝ u²
   

   ∵	s2	=	u2²	⇒	s2	=		80		²	= 4
   	s1		u1²		s1			40

   
   ∴ s₂ = 4s₁ = 8m
   If  F is retarding force and s the stopping distance, then 1/2 mv² = Fs
   For same retarding force, s α v²
   

   ∴

   s2

   =

   v2

   ²

   =

   80 km/h

   ²

   = 4

   s1

   v1

   40 km/h

   
   ∴ s₂ = 4s₁ = 4 × 2 = 8m

---

2. 
   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

   
   
   

---

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.

---

---

3. 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

   
   
   

---

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.

---

4. The displacement of a particle is represented by the following equation :
   s = 3t³ + 7t² + 5t + 8where s is in metre and t in second. The acceleration of the particle at t = 1s is

1. 1. 14 m/s²

   
   
   

   2. 18 m/s²

   
   
   

   3. 32 m/s²

   
   
   

   4. zero

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Displacement
   s = 3t³ + 7t² + 5t +8;
   Velocity = ds/dt = 9t² + 14t + 5
   

   Acceleration =	d²s	= 18t + 14
   	dt2

   
   Acceleration at (t = 1s) 
   = 18 × 1 + 14 = 18 + 14 = 32 m/s²

   Correct Option: C

   Displacement
   s = 3t³ + 7t² + 5t +8;
   Velocity = ds/dt = 9t² + 14t + 5
   

   Acceleration =	d²s	= 18t + 14
   	dt2

   
   Acceleration at (t = 1s) 
   = 18 × 1 + 14 = 18 + 14 = 32 m/s²

---

---

5. A particle moves a distance x in time t according to equation x = (t + 5)^–1. The acceleration of particle is proportional to:

1. 1. (velocity)^3/2

   
   
   

   2. (distance)²

   
   
   

   3. (distance)^–2

   
   
   

   4. (velocity)^2/3

   
   
   

---

1. View Hint View Answer Discuss in Forum

   x =	1
   	t + 5

   
   

   ∴ v =	dx	=	- 1
   	dt		(t - 5)²

   
   

   ∴ a =	d²x	=	2	= 2x³
   	dt²		(t - 5)³

   
   

   Now	1	∝ v1/2
   	(t + 5)

   
   

   ∴	1	∝ v3/2 ∝ a
   	(t + 5)³

   
   

   Correct Option: A

   x =	1
   	t + 5

   
   

   ∴ v =	dx	=	- 1
   	dt		(t - 5)²

   
   

   ∴ a =	d²x	=	2	= 2x³
   	dt²		(t - 5)³

   
   

   Now	1	∝ v1/2
   	(t + 5)

   
   

   ∴	1	∝ v3/2 ∝ a
   	(t + 5)³

   
   

---