Class 9 Science Chapter 8 Question Answers - Force and Laws of Motion

Q1: An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Ans: 

• Yes, it is possible for an object to be traveling with a non-zero velocity if it experiences a net zero external unbalanced force. This is based on Newton’s First law of motion, which states that an object will change its state of motion if and only if the net force acting on it is non-zero.  
• Thus, in order to change its motion or to bring about motion, some external unbalanced force is required. 
• In this case, the object experiences a net zero unbalanced force, which will cause it to move with some non-zero velocity provided that the object was previously moving with a constant velocity.

Q2: When a carpet is beaten with a stick, dust comes out of it. Explain.
Ans: When a carpet is beaten with a stick, dust comes out of it because of Newton’s First Law of Motion, the law of Inertia. Initially, the dust particles and the carpet are in a state of rest. When the carpet is beaten with a stick, it causes the carpet to move, while the dust particles, due to inertia of rest, will resist the change in motion. Thus, the carpet’s forward motion will exert a backward force on the dust particles, which makes them move in the opposite direction. Therefore the dust comes out of the carpet.

Q3: Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans: 

• It is advised to tie any luggage kept on the roof of a bus with a rope because of Newton’s First Law of Motion, the law of Inertia. 
• When the bus moves, the luggage will be in inertia of motion and say the bus suddenly stops, then the luggage tends to resist this change in motion, causing it to move forward and fall off, if not tied up by a rope. 
• Similarly, when the bus decelerates or changes its direction while turning, the inertia of motion of the luggage will try to resist this change in motion, causing the luggage to move oppositely and fall off, if not tied up by a rope.

Q4: A stone of 1kg is thrown with a velocity of 20ms⁻¹ across the frozen surface of a lake and comes to rest after traveling a distance of 50m. What is the force of friction between the stone and the ice?
Ans: Given:
Mass of stone: m = 1kg
Initial velocity of stone: u = 20ms⁻¹
Final velocity of stone:  v = 0ms⁻¹ (as it comes to rest)
Distance traveled on ice: s = 50m
To find: Force of friction between stone and ice.
First, we need to find the deceleration:
It is known that – v² = u² + 2as
Thus, 0² = (20)² + 2a(50)
⇒ 0  = 400 + 100a
⇒ −400 = 100a
⇒ a = −4ms⁻²
The negative sign implies deceleration.
Next, finding the frictional force:
F = ma
⇒ F = (1)(−4)
⇒ F = −4N
Thus, the force of friction between stone and ice is −4N.

Q5: An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7ms⁻²?
Ans: Given:
Mass of vehicle: m = 1500kg
Negative acceleration: a = −1.7ms⁻²
To find: Force of friction between road and vehicle.
It is known that - F = ma
⇒ F = (1500)(−1.7)
⇒ F = −2550N
Thus the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7ms⁻² is −2550N.

Q6: An object of mass 100kg is accelerated uniformly from a velocity of 5ms−1 to 8ms−1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Ans: Given:
Mass of object: m = 100kg
Initial velocity of object: u = 5ms⁻¹
Final velocity of object: v = 8ms⁻¹
Time duration of acceleration: t = 6s
To find: 

• Initial momentum
• Final momentum
• Force exerted on the object

It is known that – momentum = mass × velocity
Initial_momentum = mass × initial_velocity
⇒ Initial_momentum = 100 × 5
⇒ Initial_momentum = 500kgms⁻¹
Final_momentum = mass × final_velocity
⇒ Final_momentum = 100 × 8
⇒ Final_momentum = 800kgms⁻¹
Now, the force – F = ma

Thus,

• Initial momentum:500kgms⁻¹
• Final momentum: 800kgms⁻¹
• Force exerted on object: 50N

Q7: Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Ans: 

• Kiran’s statement – the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). 

Based on the law of conservation of momentum, the change of momentum experienced by both the insect and car should be equal. The change in velocity of the insect will be greater, due to its small mass, while the change in velocity of the car is insignificant, due to its larger mass. But the change in momentum before and after collision would be the same. Thus, Kiran’s statement is false.

• Akhtar’s statement – since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. 

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Thus both the car and insect would experience the same force. So, we can say that Akhtar’s statement is also false.

• Rahul’s statement – both the motorcar and the insect experienced the same force and a change in their momentum.

Inferring from the law of conservation of momentum and Newton’s third law of motion, we can say that Rahul’s statement is true.

Q8: State Newton’s second law of motion?
Ans: Newton’s Second law of motion states that when an unbalanced external force acts on an object, its velocity changes, that is, the object gets accelerated. It can also be stated as the time rate of change of the momentum of a body is equal in both magnitude and direction to the force applied on it.
Mathematically – F = ma, where ‘F’ is the force, ‘m’ is the mass of the object, and ‘a’ is the acceleration.

Q9:  What is the momentum of a body of mass 200g moving with a velocity of 15ms⁻¹?
Ans: Given:
Mass of body: m = 200g = 0.2kg
Velocity of body: v = 15ms⁻¹
To find: Momentum of the body.
It is known that – momentum=mass×velocity
⇒ momentum = 0.2 × 15
⇒ momentum = 3kgms−1
Thus, the momentum of the body is 3kgms⁻¹.

Q10: Define force and what are the various types of forces?
Ans: Force is defined as the push or pulls on an object that produces a change in the state or shape of the object. It can also cause a change in the speed and/or direction of motion of the object.
The various types of force are:

• Mechanical force
• Gravitational force
• Frictional force
• Electrostatic force
• Electromagnetic force
• Nuclear force

Q11: A force of 25N acts on a mass of 500g resting on a frictionless surface. What is the acceleration produced?
Ans: Given:
Mass: m = 500g = 0.5kg
Force exerted: F = 25N
To find: Acceleration.
It is known that – F = ma
⇒ a = Fm
⇒ a = 25 / 0.5
⇒ a = 50ms⁻²
Thus, the acceleration produced is 50ms⁻².

Q12: State Newton’s first law of Motion?
Ans: Newton’s first law of motion is also called the law of Inertia. It states that an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. Or, an object rest will continue to be at rest and an object in motion will continue to be in motion until and unless it is acted upon by an external force.

Q13: A body of mass 5kg starts and rolls down 32m of an inclined plane in 4s. Find the force acting on the body?
Ans: Given:
Mass of body: m = 5kg
Initial velocity of body: u = 0ms⁻¹ (as it is starting to roll)
Distance travelled on inclined plane: s = 32m
Time duration of rolling: t = 4s
To find: Force acting on the body.
First we need to find the acceleration:
It is known that

Next, finding the force:
F = ma
⇒ F = (5 × 4)
⇒ F = 20N
Thus, the force acting on the body is 20N.

Q14: On a certain planet, a small stone tossed up at 15ms⁻¹ vertically upwards takes 7.5s to return to the ground. What is the acceleration due to gravity on the planet?
Ans: Given:
Initial velocity of stone:u = 15ms⁻¹
Final velocity of stone: v = 0ms⁻¹ (as it becomes zero at the highest point)
Total time duration of flight (tossed up and falling down to the ground): t = 7.5s
To find: Acceleration due to gravity of the planet.
It is known that – v = u + at
Thus, 0 = 15 + (at)

  this denotes the time for one-half of the entire flight.
Thus the total duration of the flight is twice this duration.

Now, the acceleration due to gravity is –

Thus, the acceleration due to gravity of the planet is −4ms⁻²

Q15: Why does the passenger sitting in a moving bus are pushed in the forward direction when the bus stops suddenly?
Ans: The passengers sitting in the moving bus are pushed in the forward direction when the bus stops suddenly due to inertia because the passengers' upper body continues to be in a state of motion, while the lower part of the body that is in contact with the seat remains at rest. As a result, the passenger’s upper body is pushed in the forward direction, in the direction in which the bus was moving before coming to a halt.

Q16: Why does the boat move backward when the sailor jumps in the forward direction?
Ans: The boat moves backward when the sailor jumps in the forward direction because of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. Thus, when the sailor jumps in the forward direction he is causing an action force due to which the boat moves backward. In response, the boat exerts an equal and opposite force (the reaction force) on the sailor due to which he is pushed in the forward direction.

Q17: An astronaut has 80kg mass on earth.
(i) What is his weight on earth?
Ans: Given:
Mass of astronaut: m = 80kg
To find his weight on earth.
It is known that,
Acceleration due to gravity on earth:  g_e = 10ms⁻²
Acceleration due to gravity on mars: g_m = 3.7ms⁻²
Weight: w = m × g
Weight on earth: w_e = m × g_e
⇒ w_e =  80 × 10
w_e = 800N
(ii) What will be his mass and weight on mars with gm=3.7ms⁻²?
Ans: Given:
Mass of astronaut: m = 80kg
To find his mass and weight on mars.
It is known that,
Acceleration due to gravity on earth: g_e = 10ms⁻²
Acceleration due to gravity on mars: g_m = 3.7ms⁻²
Weight: w = m × g
Weight on mars: wm=m×gm
⇒ w_m = 80 × 3.7
w_m = 296N
The mass of astronauts remains the same on mars because it is a constant value.
Thus, mass on mars is m = 80kg.

Q18: Which of the following has more inertia:
(a) A rubber ball and a stone of the same size?
Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a rubber ball and a stone of the same size, it is clear that the stone will have greater inertia than the ball. It is because, despite being the same size, the stone weighs more than the rubber ball.
(b) A bicycle and a train?
Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a bicycle and a train, it is clear that the train will have greater inertia than the bicycle because the train weighs more than the bicycle.
(c) A five rupees coin and a one-rupee coin?
Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a five rupees coin and a one-rupee coin, the five rupees coin will have greater inertia than the one-rupee coin because five rupees coin weighs more than a one-rupee coin.

Q19: In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also, identify the agent supplying the force in each case.

Ans: In the example, the number of times the velocity of football changes is four.

(i) The velocity of the football changes first when a player kicks the ball towards another player on his team. Here, the agent supplying the force is the foot of the football player who is kicking the ball.
(ii) The velocity of the football changes second when that another player kicks the ball towards the goal. Here, the agent supplying the force is the foot of the other player who is now kicking the ball towards the goal.
(iii) The velocity of the football changes for the third time when the goalkeeper of the opposite team stops the football by collecting it. Here, the agent supplying the force are the hands of the goalkeeper who collects the ball.
(iv) The velocity of the football changes for the fourth time when the goalkeeper kicks it towards a player of his team. Here, the agent supplying the force is the foot of the goalkeeper who is now kicking the ball towards his teammate.

Q20: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans: Some of the leaves may get detached from a tree if we vigorously shake its branch because the branches of the tree will come into motion while the leaves tend to continue in their state of rest. This is due to the inertia of rest of the leaves. The force of shaking will act on the leaves with the change in direction rapidly, which results in the leaves detaching and falling off from the tree.

Q21: Why do you fall in the forward direction when a moving bus breaks to a stop and fall back when it accelerates from rest?
Ans: The passengers sitting in the moving bus are pushed in the forward direction when the bus stops suddenly due to inertia because the passengers' upper body continues to be in a state of motion, while the lower part of the body that is in contact with the seat remains at rest. As a result, the passenger’s upper body is pushed in the forward direction, in the direction in which the bus was moving before coming to a halt.
Similarly, the passengers sitting in the bus are pushed in the backward direction when the bus accelerates from rest due to inertia, because the passengers’ upper body continues to be in a state of rest, while the lower part of the body that is in contact with the seat is set in motion. As a result, the passenger’s upper body is pushed in the backward direction, in the opposite to which the bus starts to move.

Q22: If action is always equal to the reaction, explain how a horse can pull a cart.
Ans: 

• In this case, we are dealing with unbalanced forces. It is true that the horse exerts an action force on the cart and experiences a reaction force from the cart. But also, the horse creates an action force on the ground over which it is walking, and experiences a reaction force from the ground.
• In pulling the cart, the action force of the horse pulling the cart is greater than the reaction force of the cart, resisting the pull. Thus the cart moves in the direction of the pull of the horse.
• In stepping on the ground, the horse creates an action force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the horse forward.
• In this was a horse can pull a cart.

Q23: Explain why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Ans: It is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity because of Newton’s third law of motion. In this case, the water being ejected out in the forward direction with great force (action) will create a backward force that results in the backward movement (reaction) of the hose pipe. As a result of this backward force and movement, it will be difficult for the fireman to hold the hose properly with stability.

Q24: From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35ms⁻¹. Calculate the initial recoil velocity of the rifle.
Ans: Given:
Mass of rifle: m₁= 4kg
Mass of bullet: m₂ = 50g = 0.05kg
Initial velocity of rifle: u₁= 0ms⁻¹ (it is stationary during firing)
Initial velocity of bullet: u₂ = 0ms⁻¹(it starts from rest, inside the barrel of the rifle)
Fired velocity of bullet: v₂ = 35ms⁻¹
To find: Recoil velocity of rifle:v₁
By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get –
m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂
Here, m₁u₁ + m₂u₂ is the total sum of momentum of rifle and bullet before firing and m₁v₁ + m₂v₂ is the total sum of momentum of rifle and bullet after firing.
Substituting the values in – m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂

⇒ v₁ = −4.375ms⁻¹ (The negative sign indicates the backward direction in which the rifle moves when it recoils)
Thus, the recoil velocity of the rifle is 4.375ms⁻¹.

Q25: An 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40000N and the track offers a friction force of 5000N, then calculate:
(a) The net accelerating force
Ans: Given:
Force exerted by engine on wagons: F = 40000N
Frictional exerted on wagons: f = 5000N
Mass of engine: me = 8000kg
Mass of each wagon: m_w=2000kg
Mass of all five wagons: m_W = 5 × m_w = 5 × 2000 = 10000kg
Mass of entire train: m_T=m_e + m_W = 8000 + 10000 = 18000kg
To find accelerating force.
Net accelerating force can be found by subtracting the frictional force from the force exerted by the engine on the wagons.
Thus, Net Accelerating Force=Force Of Engine − Frictional Force
⇒ Net Accelerating Force = F − f
⇒ Net Accelerating Force = 40000 − 5000
⇒ Net Accelerating Force = 35000N
(b) The acceleration of the train
Ans:  Given:
Force exerted by engine on wagons: F = 40000N
Frictional exerted on wagons: f = 5000N
Mass of engine: m_e = 8000kg
Mass of each wagon: m_w = 2000kg
Mass of all five wagons: m_W = 5 × m_w = 5 × 2000 = 10000kg
Mass of entire train: m_T = m_e + m_W=8000 + 10000=18000kg
To find the acceleration of the train.
It is known that – F = ma


Q26: wo objects, each of mass 1.5kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5ms−1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Ans: Given:
Mass of object 1: m₁= 1.5kg
Mass of object 2: m2 = 1.5kg
Initial velocity of object 1: u₁= 2.5ms⁻¹
Initial velocity of object 2: u₂ = −2.5ms⁻¹ (negative sign because it is moving in the opposite direction)
Mass of combined object after collision: m = m₁ + m₂ = 1.5 + 1.5 = 3kg
To find: Final velocity of the combined object after collision:v
By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get –
m_v = m₁u₁ + m₂u₂
Here, m₁u₁ + m₂u₂ is the total sum of the momentum of objects before the collision and mv is the total momentum of the combined objects after the collision.
Substituting the values in – m_v = m₁u₁ + m₂u₂
⇒ (3 × v) = (1.5 × 2.5) + (1.5 × −2.5)
⇒ (3 × v) = (3.75) + (−3.75s)
⇒ (3 × v) = 0
⇒ v = 0ms⁻¹
Thus, the velocity of the combined object after collision is 0ms⁻¹.

Q27: According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Ans: 

• According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. This pair of forces is called the action-reaction pair. 
• In the case of the massive truck parked alongside the road, the action-reaction pair is the weight of the truck exerting a force on the road in the downward direction (action), and the static friction of the road in the upward direction (reaction), which keeps the truck at rest. These two equal and opposite forces cancel out each other, which is why the truck will not move.
• For it to move, we need to apply additional external force to overcome the static friction of the road. 
• Thus, as the student explained, the truck does not move because the two opposite and equal forces of the truck and road cancel out each other is valid.

Q28: A hockey ball of mass 200g traveling at 10ms⁻¹ is struck by a hockey stick so as to return it along its original path with a velocity at 5ms⁻¹. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans: Given:
Mass of hockey ball: m = 200g = 0.2kg
Initial velocity of hockey ball: u = 10ms⁻¹
Final velocity of hockey ball: v = −5ms⁻¹ (because it moves back in its original direction)
To find: Change in momentum of hockey ball due to the force of hockey stick
ChangeOfMomentum = m_v − m_u
Here, m_u is the initial momentum of the hockey ball and m_v is the final momentum of the hockey ball.
Substituting the values in – Change Of Momentum = m_v − m_u
⇒ ChangeOfMomentum = (0.2 × −5) − (0.2 × 10)
⇒ ChangeOfMomentum = (−1) − (2)
⇒ ChangeOfMomentum = −3kgms⁻¹
Thus, the change in momentum of hockey ball due to the force of hockey stick is −3kgms⁻¹.

Q29: A bullet of mass 10g traveling horizontally with a velocity of 150ms−1 strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Ans: Given:
Mass of bullet: m = 10g = 0.01kg
Initial velocity of bullet: u = 150ms⁻¹
Final velocity of bullet: v = 0ms⁻¹ (as it comes to rest after penetration)
Time duration of bullet travel: t = 0.03s
To find:

• Distance of penetration of bullet into the block
• Force exerted by the block on the bullet

(a) Distance of penetration:
It is known that – v = u + at
Thus, 0 = 150 + (a × 0.03)
⇒ (a × 0.03) = −150
⇒ a = −5000ms⁻²
Now,

(b) Next, finding the force:
F = ma
⇒ F = (0.01 × −5000)
⇒ F = −50N
Thus,

• Distance of penetration of bullet into the block is 2.25m
• Force exerted by the block on the bullet is −50N

Q30: Differentiate between mass and weight?
Ans: The difference between mass and weight is given below,


Q31: A scooter is moving with a velocity of 20ms−1when brakes are applied. The mass of the scooter and the rider is 180kg. The constant force applied by the brakes is 500N.
(a) How long should the brakes be applied to make the scooter comes to a halt?
Ans:  Given:
Mass of scooter and rider: m = 180kg
Initial velocity of scooter: u = 20ms−1
Final velocity of scooter: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −500N(as it opposes the motion)
To find the time duration over which brake should be applied to stop the scooter.
Now, the force – F = ma

Rearranging,

(b) How far does the scooter travel before it comes to rest?
Ans: Given:
Mass of scooter and rider: m = 180kg
Initial velocity of scooter: u = 20ms⁻¹
Final velocity of scooter: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −500N(as it opposes the motion)
To find distance travelled by scooter before coming to halt.

Acceleration is negative because it is retarding the motion of the scooter.

Q32: State Newton’s third law of motion and how does it explain the walking of man on the ground?
Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction acting on different bodies. This implies the existence of the action-reaction force pair. That is, for every action force created an equal and opposite reaction force will be created.
The walking of a man on the ground can be explained with Newton’s third law of motion. During walking on the ground, the man creates an active force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the man forward enabling him to walk.

Q33: State Newton’s second law of motion and derive it mathematically?
Ans: Newton’s second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Mathematical derivation:
Say we have an object of mass m that is moving along a straight line with an initial velocity, u.
It is then uniformly accelerated to velocity, v in time, t  by the application of a constant force, F throughout the time, t.
Initial Momentum of object: p₁= m × u
Final Momentum of object: p₂= m × v
Now, the change of momentum is the Final momentum subtracted by the Initial momentum
Thus, Δp = p₂ − p₁= m_v − m_u = m(v − u)
⇒ Δp = m(v − u)
The rate of change of momentum is Δp / t

We know that the applied force is proportional to the rate of change of momentum of the object.

Using these, we get
⇒ F = ma
The SI unit of force is Newton (Kgms⁻²)
The second law of motion gives a method to measure the force acting on an object as a product of its mass and acceleration.

Q34: Why does a person while firing a bullet holds the gun tightly to his shoulders?
Ans: A person while firing a bullet holds the gun tightly to his shoulder because of the recoil of the gun when the bullet is fired. This is in accordance with Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction, that acts on different bodies. 

• When a bullet is fired, the forward motion of the bullet (action) creates a recoil or backward motion of the gun (reaction). The action force being much greater will create an equivalent recoil force in the backward direction. 
• If the person who holds the gun does not hold it properly to his shoulders that can result in injury. This is because the shoulder absorbs most of the force during recoil that enables the shooter to take a steady shot.
• Thus, if not held tightly to his shoulders, the shot will not be precise and this can also cause the gun to fly away from his hands.

Q35: A car is moving with a velocity of 16ms−1 when brakes are applied. The force applied by the brakes is 1000N. The mass of the car its passengers is 1200kg.
(a) How long should the brakes be applied to make the car come to a halt?
Ans: Given:
Mass of car and passengers: m = 1200kg
Initial velocity of car: u = 16ms⁻¹
Final velocity of car: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −1000N(as it opposes the motion)
To find time duration over which brake should be applied to stop the car.
Now, the force – F = ma

Rearranging,

(b) How far does the car travel before it comes to rest?
Ans: Given:
Mass of car and passengers: m = 1200kg
Initial velocity of car: u = 16ms⁻¹
Final velocity of car: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −1000N(as it opposes the motion)
To find distance travelled by car before coming to halt.
Acceleration -

Acceleration is negative because it is retarding the motion of the scooter.