NCERT Solutions · Class 11 Physics Laws of Motion

The test in every part is the same: is the velocity changing? If not, Newton's second law forces the net force to be exactly zero.Newton's second law says \(\displaystyle \vec{F}_{net} = m\vec{a} \), where \(\displaystyle \vec{a} \) is the acceleration of the body. Whenever the acceleration is zero — whether the body is at rest or moving with constant velocity — the net force acting on it must be zero, no matter how many individual forces (gravity, tension, friction, drag, upthrust) are actually pushing and pulling on it. Those individual forces don't vanish; they simply add up to zero as vectors. A zero vector has no direction, so in each case below the answer is "zero" with no direction attached.(a) Raindrop falling with constant speed. "Falling down with a constant speed" means the velocity is not changing — it is uniform. Zero rate of change of velocity means zero acceleration: \[\vec{a} = 0 \implies \vec{F}_{net} = m\vec{a} = 0 \] Physically, this happens because the drop's weight \(\displaystyle mg \) acting downward is exactly balanced by the upward viscous drag force of the air once the drop reaches its constant ("terminal") speed. The two are equal and opposite, so they cancel. Net force = $\displaystyle 0$ N.(b) A $\displaystyle 10$ g cork floating on water. The cork is stationary — at rest, and staying at rest — so again \(\displaystyle \vec{a} = 0 \). The mass of $\displaystyle 10$ g is not needed for the force calculation; it would only matter if the cork were accelerating. The weight of the cork, \(\displaystyle mg \), acts downward, and it is balanced by an equal upward upthrust (buoyant force) from the water it displaces. Net force = $\displaystyle 0$ N.(c) A kite held stationary in the sky. "Stationary" again means \(\displaystyle \vec{a} = 0 \), so \(\displaystyle \vec{F}_{net} = 0 \). Here three effects are at work — the kite's weight downward, the tension in the string, and the net aerodynamic force from the wind — and a skilled flier holds the kite so that these add up, as vectors, to zero. Net force = $\displaystyle 0$ N.(d) A car moving at a constant $\displaystyle 30$ km/h on a rough road. The key word is constant velocity — this is uniform motion, not acceleration, so once more \(\displaystyle \vec{a} = 0 \) and \(\displaystyle \vec{F}_{net} = 0 \). This is the case people most often get wrong: a car needs its engine running and burning fuel, so it feels like there "must" be a net force pushing it forward. But the engine's forward driving force on the wheels is exactly canceled by the backward resistive forces — friction from the rough road and air resistance — because the speed isn't changing. The $\displaystyle 30$ km/h value itself never enters the calculation; only the fact that it is constant matters. Net force = $\displaystyle 0$ N.(e) A high-speed electron far from all matter, with no electric or magnetic field. This case is different in kind from the other four. In (a)–(d) several real forces exist and happen to cancel. Here, the electron is stated to be far from all material objects (so gravitational pull on it is negligible) and explicitly free of electric and magnetic fields (so there is no electromagnetic force either). With no source of any force acting on it at all, there is nothing to sum — the net force is zero because zero forces act, not because opposing forces balance. By Newton's first law, it simply continues moving in a straight line at its constant high speed. Net force = $\displaystyle 0$ N.Answer: In every case the net force is zero — (a) raindrop: drag balances weight; (b) cork: upthrust balances weight; (c) kite: string tension and wind force balance weight; (d) car: resistive forces balance the engine's driving force; (e) electron: no force acts on it at all. None of the five has a direction, since the net force in each is the zero vector.

The only force on the pebble in flight is gravity — and gravity does not care which way the pebble is moving.Once the pebble leaves the hand, air resistance is ignored, so the pebble is in free fall: the sole force acting on it is its own weight, which by Newton's second law and the definition of weight is\[F = mg \]where \(\displaystyle m\) is the mass of the pebble and \(\displaystyle g\) is the acceleration due to gravity. This force is the pebble's weight, not its mass — mass ($\displaystyle 0.05$ kg) is a property of the pebble itself and stays the same throughout the flight, while weight is the force gravity exerts on that mass, and it is weight that determines the net force here.Substituting \(\displaystyle m = 0.05\ \text{kg}\) and \(\displaystyle g = 10\ \text{m s}^{-2}\) (as instructed in the problem):\[F = 0.05\ \text{kg} \times 10\ \text{m s}^{-2} = 0.5\ \text{N} \]This force acts vertically downward (toward the Earth), because weight always points toward the Earth's centre — it never reverses with the object's motion.(a) During upward motion: The pebble is moving up, but the force on it is still just its weight. So the net force is $\displaystyle 0.5$ N, directed vertically downward. (This is exactly why the pebble decelerates as it rises — the force opposes the velocity.)(b) During downward motion, and at the highest point: Nothing about the source of the force has changed — gravity is still the only force acting, unopposed by air resistance. So the net force is again $\displaystyle 0.5$ N, directed vertically downward. At the highest point, the pebble's velocity is momentarily zero, but its acceleration is not: a common mistake is to think "zero velocity means zero force," but velocity and force are independent quantities here. The net force at the highest point is still exactly \(\displaystyle mg = 0.5\ \text{N}\) downward, which is precisely why the pebble does not stay at rest but immediately begins to fall.Does throwing it at $\displaystyle 45$° change anything? No. If the pebble is launched at $\displaystyle 45$° to the horizontal, its velocity has both a horizontal and a vertical component, and it follows a curved (projectile) path instead of a straight vertical line. But the force analysis is identical: with air resistance ignored, gravity is still the only force acting at every instant of the flight, and gravity acts only vertically — it has no horizontal component and does not depend on the direction of the pebble's velocity. So the net force is still \(\displaystyle mg = 0.5\ \text{N}\), vertically downward, throughout the flight — during the rise, at the top of the arc, and during the fall — exactly as in the straight-up case. The angle of projection changes the trajectory (and how the velocity components evolve), not the force.Answer: In every case — rising, falling, and at the highest point, whether thrown straight up or at $\displaystyle 45$° — the net force on the pebble is $\displaystyle 0.5$ N, directed vertically downward, equal to its weight \(\displaystyle mg\); the direction and magnitude never change during the flight.

Once a stone leaves your hand, only forces that are still touching it can act on it — for a falling stone that force is gravity alone, no matter what the train was doing at the instant you let go.Two ideas do all the work here: Newton's second law, \(\displaystyle F = ma\), where \(\displaystyle m\) is the body's mass and \(\displaystyle a\) its acceleration, and Newton's first law — a body needs no force at all to keep moving at whatever velocity it already has, only to change that velocity. Weight is the special case of the second law with \(\displaystyle a = g\): \(\displaystyle W = mg\).Given: \(\displaystyle m = 0.1\) kg, and for this problem \(\displaystyle g = 10\) m \(\displaystyle s^{-2}\).(a) Dropped from a window of a stationary trainThe instant the stone leaves your hand it is touching nothing but air, which is neglected here. The only force left on it is gravity.\[F = mg = 0.1 \times 10 = 1\ \text{N}\]Net force: $\displaystyle 1$ N, vertically downward.(b) Dropped from a train running at a constant $\displaystyle 36$ km/hConverting only to see why it doesn't matter: \(\displaystyle 36\ \text{km/h} = 36 \times \dfrac{5}{18} = 10\) m \(\displaystyle s^{-1}\). A train moving at constant velocity is itself an inertial frame, so by Newton's first law nothing was pushing the stone horizontally even before it was dropped — it was simply coasting along with the train, force-free, at $\displaystyle 10$ m \(\displaystyle s^{-1}\). Releasing it changes nothing about that: the horizontal velocity persists on its own (no force needed to sustain a constant velocity) while, exactly as in (a), gravity is the only actual force now acting.\[F = mg = 0.1 \times 10 = 1\ \text{N}\]Net force: $\displaystyle 1$ N, vertically downward — the same as case (a). A constant horizontal velocity, however large, contributes nothing to the net force; only a change in velocity needs one.(c) Dropped from a train accelerating at $\displaystyle 1$ m \(\displaystyle s^{-2}\)This is the one place it's tempting to add in a horizontal "acceleration force," and that would be the mistake. While the stone was still in your hand or on the sill, the train's frame (through your grip) was indeed pushing it forward at $\displaystyle 1$ m \(\displaystyle s^{-2}\). But "just after it is dropped" means contact is already gone — and a force can only act through contact (or gravity, which needs none). From that instant the stone is a free body exactly like in (a) and (b): whatever horizontal velocity it happened to have when released (equal to the train's velocity at that moment) simply continues unchanged by Newton's first law, needing no force to persist, while the one real force still acting is gravity.\[F = mg = 0.1 \times 10 = 1\ \text{N}\]Net force: $\displaystyle 1$ N, vertically downward — identical to (a) and (b). The train's acceleration before release, and the stone's leftover horizontal velocity after it, never show up in the net force once the stone is falling freely.(d) Lying on the floor of a train accelerating at $\displaystyle 1$ m \(\displaystyle s^{-2}\), at rest relative to the trainHere the stone never loses contact — it rides along with the floor, so it shares the train's acceleration itself: \(\displaystyle a = 1\) m \(\displaystyle s^{-2}\), horizontal, in the train's direction of motion.\[F = ma = 0.1 \times 1 = 0.1\ \text{N}\]This $\displaystyle 0.1$ N is supplied by friction between the stone and the floor; friction is what drags the stone forward with the accelerating floor instead of letting it slide backward relative to the train.Net force: $\displaystyle 0.1$ N, horizontal, in the direction of the train's acceleration.The contrast between (c) and (d) is the point of the question: the same accelerating train gives two different answers depending on whether the stone is still in contact with it. Off the train and falling, only gravity acts, however the train was moving. On the train's floor, the stone is forced to share the train's acceleration, and it is friction — not gravity — that supplies that force.Answer: (a) $\displaystyle 1$ N, vertically downward. (b) $\displaystyle 1$ N, vertically downward. (c) $\displaystyle 1$ N, vertically downward. (d) $\displaystyle 0.1$ N, horizontal, in the direction of the train's acceleration.

Only one horizontal force acts on the particle — the string tension — so the net centripetal force is just \(\displaystyle T\), not \(\displaystyle T\) plus or minus something else.A common trap in this question is to treat \(\displaystyle \dfrac{mv^2}{l}\) as if it were a separate, real force sitting alongside the tension, so that the net force becomes \(\displaystyle T - \dfrac{mv^2}{l}\) or \(\displaystyle T + \dfrac{mv^2}{l}\). It is not a force at all — it is the value that the net centripetal force must equal for circular motion to happen. Confusing "the quantity \(\displaystyle mv^2/l\)" with "a force called \(\displaystyle mv^2/l\)" is exactly what options (ii) and (iii) are testing.Step $\displaystyle 1$: List every force on the particle.The particle of mass \(\displaystyle m\) sits on a horizontal table and is tied by a string to a fixed peg on that table. Three forces act on it:1. Gravity, \(\displaystyle mg\), vertically downward. 2. The normal reaction \(\displaystyle N\) from the table, vertically upward. 3. The tension \(\displaystyle T\) in the string, horizontal, directed from the particle toward the peg (i.e., toward the centre of the circle).The table is smooth, so there is no friction — no horizontal force from the table surface.Step $\displaystyle 2$: Resolve vertically.The particle stays on the table (no vertical acceleration), so the vertical forces balance: \[N = mg. \] This tells us the table supplies exactly enough upward push to cancel weight, and nothing more. It plays no role in the horizontal (circular) motion.Step $\displaystyle 3$: Resolve horizontally — this is the circular-motion direction.The only horizontal force on the particle is the tension \(\displaystyle T\), and it points toward the centre of the circle (that is what a taut string tied to a fixed peg does). Since it is the only horizontal force, it is automatically the net force in that direction: \[F_{\text{net, toward centre}} = T. \]Step $\displaystyle 4$: Check this against Newton's second law for circular motion.For any particle moving in a circle of radius \(\displaystyle l\) with speed \(\displaystyle v\), Newton's second law requires the net force directed toward the centre to equal the centripetal force: \[F_{\text{net}} = \frac{mv^2}{l}. \] Combining this with Step $\displaystyle 3$ gives \[T = \frac{mv^2}{l}. \] This is a consequence — it tells you how large the tension must be for the particle to hold that circular path at that speed. It is not license to write \(\displaystyle T - \dfrac{mv^2}{l}\) or \(\displaystyle T + \dfrac{mv^2}{l}\) as "the net force": doing so would be adding the same quantity to itself under a different name, since \(\displaystyle T\) and \(\displaystyle mv^2/l\) are numerically identical here, not two separate forces to be combined.Step $\displaystyle 5$: Eliminate the wrong options explicitly. Only option (i) is consistent with there being exactly one horizontal force — the tension — supplying the entire centripetal requirement.Answer: (i) T — the string tension alone is the net force toward the centre; it satisfies \(\displaystyle T = mv^2/l\), but that equation is a consequence of Newton's second law, not a second force to add or subtract.

A retarding force means the force acts opposite to the direction of motion, so it produces a deceleration — the object slows down at a steady rate until its velocity reaches zero.Step $\displaystyle 1$: Find the deceleration using Newton's second law.Newton's second law relates the net force \(\displaystyle F \) on a body to the mass \(\displaystyle m \) and the acceleration \(\displaystyle a \) it produces: \[F = ma \]Here the body has mass \(\displaystyle m = 20 \text{ kg} \), and a retarding force of magnitude \(\displaystyle F = 50 \text{ N} \) acts on it. Since the force opposes the motion, it produces a deceleration rather than an acceleration — the magnitude of this deceleration is\[a = \frac{F}{m} = \frac{50 \text{ N}}{20 \text{ kg}} = 2.5 \text{ m s}^{-2} \]directed opposite to the body's velocity.Step $\displaystyle 2$: Use the equation of motion to find the time to stop.Take the direction of the initial velocity as positive. Then the retarding force gives an acceleration of \(\displaystyle -2.5 \text{ m s}^{-2} \) (negative because it opposes the motion). The initial speed is \(\displaystyle u = 15 \text{ m s}^{-1} \), and the body "stops" when its final velocity \(\displaystyle v = 0 \).Using the first equation of motion, \[v = u + at \]substitute the known values: \[0 = 15 \text{ m s}^{-1} + (-2.5 \text{ m s}^{-2}) \, t \]\[2.5 \, t = 15 \]\[t = \frac{15 \text{ m s}^{-1}}{2.5 \text{ m s}^{-2}} = 6 \text{ s} \]A common slip here is to keep the sign of \(\displaystyle a\) positive and get a negative time — always assign the retarding force a negative sign relative to the chosen direction of motion, since it acts to reduce the speed, not increase it.The given quantities ($\displaystyle 50$ N, $\displaystyle 20$ kg, $\displaystyle 15$ m s⁻¹) each carry two significant figures, and the division \(\displaystyle \frac{15}{2.5} \) comes out to exactly $\displaystyle 6$, so no further rounding is needed.Answer: The body takes $\displaystyle 6$ s to stop.

A change in speed over a time interval, with no change in direction, means the acceleration is uniform and along the line of motion — so \(\displaystyle v = u + at \) applies directly.The body moves in a straight line throughout (the problem says the direction is unchanged), so there is no need to treat this as a vector problem in two dimensions — one axis, taken along the direction of motion, is enough.Step $\displaystyle 1$: Find the acceleration.From the first equation of motion, \[v = u + at \] where \(\displaystyle u = 2.0 \text{ m s}^{-1} \) is the initial speed, \(\displaystyle v = 3.5 \text{ m s}^{-1} \) is the final speed, and \(\displaystyle t = 25 \text{ s} \) is the time taken. Solving for \(\displaystyle a \), \[a = \frac{v - u}{t} = \frac{3.5 \text{ m s}^{-1} - 2.0 \text{ m s}^{-1}}{25 \text{ s}} = \frac{1.5 \text{ m s}^{-1}}{25 \text{ s}} = 0.06 \text{ m s}^{-2} \]Step $\displaystyle 2$: Find the force from Newton's second law.Newton's second law states \(\displaystyle F = ma \), where \(\displaystyle m = 3.0 \text{ kg} \) is the mass of the body. Substituting, \[F = ma = 3.0 \text{ kg} \times 0.06 \text{ m s}^{-2} = 0.18 \text{ N} \]Both \(\displaystyle u \), \(\displaystyle v \), \(\displaystyle m \), and \(\displaystyle t \) are given to two significant figures, so the answer is kept to two significant figures: \(\displaystyle 0.18 \text{ N} \) (no further rounding is needed — the calculation already lands there).Step $\displaystyle 3$: Find the direction.The speed increases from $\displaystyle 2.0$ m s⁻¹ to $\displaystyle 3.5$ m s⁻¹, so the acceleration — and hence the force, since \(\displaystyle F = ma \) with \(\displaystyle m > 0 \) — points the same way the body is already moving. A force that speeds a body up (rather than slowing it down) always acts along the direction of motion, not against it; only a decelerating force would point opposite to the velocity.Answer: The force has magnitude $\displaystyle 0.18$ N, and it acts in the direction of motion of the body.

Two forces at right angles combine like the legs of a right triangle — their resultant is the hypotenuse.The body has only these two forces acting on it, so Newton's second law, \(\displaystyle \vec{F}_{net} = m\vec{a} \), needs the resultant of the $\displaystyle 8$ N and $\displaystyle 6$ N forces, not their sum or difference.Step $\displaystyle 1$: Find the resultant force.Since the two forces \(\displaystyle F_1 = 8\ \text{N} \) and \(\displaystyle F_2 = 6\ \text{N} \) are perpendicular, they add like vectors at $\displaystyle 90$°, so the Pythagorean theorem gives the magnitude of the resultant \(\displaystyle F \) directly:\[F = \sqrt{F_1^2 + F_2^2} = \sqrt{(8\ \text{N})^2 + (6\ \text{N})^2} = \sqrt{64 + 36}\ \text{N} = \sqrt{100}\ \text{N} = 10\ \text{N} \]Step $\displaystyle 2$: Find the acceleration from Newton's second law.Newton's second law states \(\displaystyle F = ma \), where \(\displaystyle m \) is the mass and \(\displaystyle a \) is the acceleration produced. Here \(\displaystyle m = 5\ \text{kg} \) and \(\displaystyle F = 10\ \text{N} \), so\[a = \frac{F}{m} = \frac{10\ \text{N}}{5\ \text{kg}} = 2\ \text{m s}^{-2} \]Both given forces ($\displaystyle 8$ N, $\displaystyle 6$ N) and the mass ($\displaystyle 5$ kg) have one significant figure quoted as whole numbers with no trailing uncertainty implied beyond that, so the result is reported to the same precision: \(\displaystyle 2\ \text{m s}^{-2} \), not \(\displaystyle 2.0 \) or \(\displaystyle 2.00\ \text{m s}^{-2} \).Step $\displaystyle 3$: Find the direction.An acceleration is a vector — quoting only "$\displaystyle 2$ m s⁻²" is incomplete. The acceleration points along the resultant force, so find the angle \(\displaystyle \theta \) that the resultant makes with the $\displaystyle 8$ N force using\[\tan\theta = \frac{F_2}{F_1} = \frac{6\ \text{N}}{8\ \text{N}} = 0.75 \]\[\theta = \tan^{-1}(0.75) = 37^\circ \](Take care here: \(\displaystyle \theta \) is measured from the $\displaystyle 8$ N force toward the $\displaystyle 6$ N force — swapping which force is in the numerator would instead give the angle from the $\displaystyle 6$ N force, \(\displaystyle 53^\circ \).)Answer: The acceleration has magnitude \(\displaystyle 2\ \text{m s}^{-2} \), directed along the resultant of the two forces, making an angle of \(\displaystyle 37^\circ \) with the $\displaystyle 8$ N force (equivalently, \(\displaystyle 53^\circ \) with the $\displaystyle 6$ N force).

The vehicle plus driver together form one system — Newton's second law needs the total mass being decelerated, not just the three-wheeler's mass.Step $\displaystyle 1$: Convert the initial speed to SI units.\[u = 36\ \text{km/h} = 36 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 10\ \text{m s}^{-1} \]Step $\displaystyle 2$: Find the acceleration (retardation) from the given time.The three-wheeler comes to rest, so the final velocity \(\displaystyle v = 0\), in a time \(\displaystyle t = 4.0\ \text{s}\). Using the first equation of motion \(\displaystyle v = u + at\):\[a = \frac{v-u}{t} = \frac{0 - 10\ \text{m s}^{-1}}{4.0\ \text{s}} = -2.5\ \text{m s}^{-2} \]The negative sign shows this is a retardation (a deceleration) — the acceleration points opposite to the direction of motion.Step $\displaystyle 3$: Identify the total mass being decelerated.A common slip here is to use only the vehicle's mass. The driver sits on the vehicle and decelerates along with it, so the driver's mass is part of the system that Newton's second law acts on:\[m = m_{\text{vehicle}} + m_{\text{driver}} = 400\ \text{kg} + 65\ \text{kg} = 465\ \text{kg} \]Step $\displaystyle 4$: Apply Newton's second law, \(\displaystyle F = ma\).Here \(\displaystyle F\) is the net (retarding) force, \(\displaystyle m = 465\ \text{kg}\) is the total mass, and \(\displaystyle a = -2.5\ \text{m s}^{-2}\) is the acceleration found above.\[F = m a = 465\ \text{kg} \times (-2.5\ \text{m s}^{-2}) = -1162.5\ \text{N} \]The negative sign again just means the force acts backward, opposite to the vehicle's direction of travel — that is what makes it a retarding (braking) force. Its magnitude is \(\displaystyle 1162.5\ \text{N}\).Step $\displaystyle 5$: Round to the correct number of significant figures.The speed ($\displaystyle 36$ km/h) and the time ($\displaystyle 4.0$ s) are each given to $\displaystyle 2$ significant figures, so the acceleration and hence the force cannot be trusted beyond $\displaystyle 2$ significant figures either — carrying the raw multiplication ($\displaystyle 1162.5$ N) as the final answer would overstate the precision of the data. Rounding once, at the end:\[F \approx 1.2 \times 10^{3}\ \text{N} = 1200\ \text{N} \]This retarding force acts on the vehicle in the direction opposite to its motion (i.e., backward, the direction of braking).Answer: The average retarding force on the vehicle is about \(\displaystyle 1.2 \times 10^{3}\ \text{N}\) ($\displaystyle 1200$ N), directed opposite to the vehicle's motion.

The rocket needs an upward force big enough to cancel its weight and still leave a net force for the acceleration — that force is the thrust.Take upward as positive. Two forces act on the rocket during lift-off: the thrust \(\displaystyle F \) from the blast (upward) and its weight \(\displaystyle mg \) (downward, always present, everywhere). By Newton's second law, the net upward force must equal mass times the given acceleration \(\displaystyle a \):\[F - mg = ma \]So the thrust itself is\[F = m(g + a) \]This is the step people skip: thrust is not just \(\displaystyle ma \). Part of the engine's force is used purely to hold the rocket's weight up; only the leftover produces the actual acceleration. Forgetting the \(\displaystyle mg \) term is the classic mistake here.Given: Substitute:\[F = (2.0 \times 10^{4} \text{ kg})(10 \text{ m s}^{-2} + 5.0 \text{ m s}^{-2}) \]\[F = (2.0 \times 10^{4} \text{ kg})(15 \text{ m s}^{-2}) \]\[F = 3.0 \times 10^{5} \text{ N} \]Both \(\displaystyle a \) and \(\displaystyle g \) are given to two significant figures, so the final thrust is quoted to two significant figures as well: \(\displaystyle 3.0 \times 10^{5} \text{ N} \), not \(\displaystyle 300000 \text{ N} \) written out, which would falsely suggest six-figure precision.The direction of this thrust is straight upward, the same direction as the acceleration — it has to be, since it must both support the weight and accelerate the rocket further up.Answer: The initial thrust of the blast is \(\displaystyle 3.0 \times 10^{5} \text{ N} \), directed vertically upward.

A single "one-formula" answer is the trap here — the body only accelerates while the $\displaystyle 8.0$ N force is actually pushing on it. Before \(\displaystyle t=0\) and after the force switches off at \(\displaystyle t=30\ \text{s}\), it coasts at constant velocity, so the problem splits into three separate pieces of motion.Take north as the positive \(\displaystyle x\)-direction (the direction of the initial velocity), so south is negative. Then: Acceleration while the force acts. By Newton's second law, \(\displaystyle F = ma\): \[a = \frac{F}{m} = \frac{-8.0\ \text{N}}{0.40\ \text{kg}} = -20\ \text{m s}^{-2} \] The minus sign means the acceleration points south — it first slows the northward motion, then reverses it. This acceleration exists only for \(\displaystyle 0 \le t \le 30\ \text{s}\); outside that window \(\displaystyle a = 0\).(a) At \(\displaystyle t = -5\ \text{s}\). This is before the force is applied, so the body still moves at its original constant velocity \(\displaystyle u = +10\ \text{m s}^{-1}\). With \(\displaystyle x(0)=0\) as the reference, \[x(-5) = u\,t = (10\ \text{m s}^{-1})(-5\ \text{s}) = -50\ \text{m} \] Negative means the body was \(\displaystyle 50\ \text{m}\) south of the origin five seconds earlier, still travelling north toward it — exactly what a constant \(\displaystyle 10\ \text{m s}^{-1}\) northward velocity requires.(b) At \(\displaystyle t = 25\ \text{s}\). This lies inside \(\displaystyle 0 \le t \le 30\ \text{s}\), so the force is still acting and the accelerated formula applies directly: \[x(t) = x(0) + u t + \tfrac12 a t^2 \] \[x(25) = 0 + (10\ \text{m s}^{-1})(25\ \text{s}) + \tfrac12(-20\ \text{m s}^{-2})(25\ \text{s})^2 \] \[x(25) = 250\ \text{m} - (10)(625)\ \text{m} = 250\ \text{m} - 6250\ \text{m} = -6000\ \text{m} \] So \(\displaystyle x(25\ \text{s}) = -6.0\times10^{3}\ \text{m}\) — \(\displaystyle 6.0\ \text{km}\) south of the origin.(c) At \(\displaystyle t = 100\ \text{s}\). The force has already switched off at \(\displaystyle t=30\ \text{s}\), so plugging \(\displaystyle t=100\ \text{s}\) straight into the accelerated formula would be wrong — the acceleration was never present past \(\displaystyle 30\ \text{s}\). Instead, find the velocity and position at the moment the force stops, then let the body coast.Velocity at \(\displaystyle t=30\ \text{s}\): \[v(30) = u + a t = 10\ \text{m s}^{-1} + (-20\ \text{m s}^{-2})(30\ \text{s}) = 10 - 600 = -590\ \text{m s}^{-1} \] Position at \(\displaystyle t=30\ \text{s}\): \[x(30) = 0 + (10)(30) + \tfrac12(-20)(30)^2 = 300\ \text{m} - 9000\ \text{m} = -8700\ \text{m} \] For the remaining interval from \(\displaystyle t=30\ \text{s}\) to \(\displaystyle t=100\ \text{s}\) (a duration of \(\displaystyle 70\ \text{s}\)) there is no force, so the velocity stays fixed at \(\displaystyle -590\ \text{m s}^{-1}\): \[x(100) = x(30) + v(30)\,(100-30) = -8700\ \text{m} + (-590\ \text{m s}^{-1})(70\ \text{s}) \] \[x(100) = -8700\ \text{m} - 41300\ \text{m} = -50000\ \text{m} \] So \(\displaystyle x(100\ \text{s}) = -5.0\times10^{4}\ \text{m}\) — \(\displaystyle 50\ \text{km}\) south of the origin.Significant figures. Every given quantity (\(\displaystyle 0.40\ \text{kg}\), \(\displaystyle 10\ \text{m s}^{-1}\), \(\displaystyle 8.0\ \text{N}\)) carries two significant figures, so no intermediate rounding was done — each result above is reported to two significant figures only at the end: \(\displaystyle -50\ \text{m}\), \(\displaystyle -6.0\times10^3\ \text{m}\), \(\displaystyle -5.0\times10^4\ \text{m}\).All three positions are negative, meaning south of the point \(\displaystyle x=0\) — the body never gets back to the north side of the origin once the southward force starts overpowering it.Answer: \(\displaystyle x(-5\ \text{s}) = -50\ \text{m}\) ($\displaystyle 50$ m south of the origin); \(\displaystyle x(25\ \text{s}) = -6.0\times10^{3}\ \text{m}\) ($\displaystyle 6.0$ km south); \(\displaystyle x(100\ \text{s}) = -5.0\times10^{4}\ \text{m}\) ($\displaystyle 50$ km south) — positions measured along the north–south line with north taken positive.