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The driver of a three wheeler moving wit...

The driver of a three wheeler moving with a speed of `36 km//h` sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child What is the average retarding force on the vehicle ? The mass of three wheeler is 400 kg and mass of the driver is 65 kg.

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To solve the problem, we will follow these steps: ### Step 1: Convert the speed from km/h to m/s The initial speed of the three-wheeler is given as 36 km/h. We need to convert this speed into meters per second (m/s). \[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \] ...
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Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} If an object of mass 2 kg and constant b = 4 (N-s)/(m) has terminal speed v_(T) in a liquid then time required to reach 0.63 v_(T) from start of the motion is :

Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil. The sphere approaches a terminal speed of 10.00 cm/s. Time required to achieve speed 6.32 cm/s from start of the motion is (Take g = 10.00 m//s^(2) ) :

Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends on the properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} Which object would first acquire half of their respective terminal speed in minimum time from start of the motion of all were released simultaneously ?

NCERT ENGLISH-LAWS OF MOTION -EXERCISE
  1. A constant force acting on a body of mass 3 kg changes its speed from ...

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  2. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and ...

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  3. The driver of a three wheeler moving with a speed of 36 km//h sees a c...

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  4. A rocket with a lift off mass 20000 kg is blasted upwards with a net i...

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  5. A body of mass 0.40 kg moving initially with a constant speed of 10 m...

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  6. A truck starts from rest and accelerate uniformly with 2 ms^(-2). At t...

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  7. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m l...

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  8. A man of mass 70 kg stands on a weighing machine in a lift, which is m...

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  9. Fig. shows the position-time graph of a particle of mass 4 kg. What is...

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  10. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth hor...

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  11. Two masses 8 kg and 12 kg are connected at the two ends of a light in ...

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  12. A nucleus is at rest in the laboratory frame of reference. Show that i...

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  13. Two billiard balls each of mass 0.05 kg moving in opposite directions ...

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  14. A shell of mass 200g is fired by a gun of mass 100kg. If the muzzle sp...

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  15. A batsman deflects a ball by an angle of 45^(@) without changing its i...

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  16. A stone of of mass 0.25 kg tied to the end of a string is whirled roun...

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  17. If in Q . 21 the speed of the stone is increased beyond the maximum pe...

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  18. Explain why (a) A horse cannot pull a cart and run in empty space. ...

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  19. Figure shows the position-time graph of a particle of mass 0.04kg. Sug...

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  20. Figure shown a man standing stationary with respect to a horizontal co...

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