NCERT Solutions · Class 11 Physics Motion in a Plane

Average speed looks at how far the wheels actually turned; average velocity looks only at where you ended up relative to where you started. For a route that bends, those are two different numbers — and comparing them is exactly the point of this question.Average speed is defined as\[v_{\text{avg}} = \frac{\text{total path length}}{\text{total time taken}} \]Here the path length is the actual distance covered along the road, however curved it is — the cab's odometer reading, not the straight-line map distance.Average velocity, by contrast, is a vector defined as\[\vec{v}_{\text{avg}} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\text{displacement}}{\text{total time taken}} \]Displacement is the straight-line vector from the starting point (station) to the ending point (hotel), regardless of the path taken to get there. The question asks only for its magnitude, so no direction needs to be reported here.Before substituting, convert the time to hours so it matches the kilometre data:\[\Delta t = 28\ \text{min} = \frac{28}{60}\ \text{h} = \frac{7}{15}\ \text{h} \](A common slip is plugging minutes straight into a km/h formula — always convert to a consistent unit pair first.)(a) Average speedThe actual road path is $\displaystyle 23$ km long, so\[v_{\text{avg}} = \frac{23\ \text{km}}{\dfrac{7}{15}\ \text{h}} = 23 \times \frac{15}{7}\ \text{km/h} = \frac{345}{7}\ \text{km/h} = 49.29\ldots\ \text{km/h} \]All three given quantities ($\displaystyle 23$ km, $\displaystyle 10$ km, $\displaystyle 28$ min) carry two significant figures, so the result is rounded to two significant figures at the end:\[v_{\text{avg}} \approx 49\ \text{km/h} \](b) Magnitude of average velocityThe displacement is the straight road distance between station and hotel, $\displaystyle 10$ km — not the $\displaystyle 23$ km actually driven:\[|\vec{v}_{\text{avg}}| = \frac{10\ \text{km}}{\dfrac{7}{15}\ \text{h}} = 10 \times \frac{15}{7}\ \text{km/h} = \frac{150}{7}\ \text{km/h} = 21.43\ldots\ \text{km/h} \]Rounded to two significant figures:\[|\vec{v}_{\text{avg}}| \approx 21\ \text{km/h} \]Are the two equal? No. Average speed ($\displaystyle 49$ km/h) is larger than the magnitude of average velocity ($\displaystyle 21$ km/h), and this is not a coincidence: distance travelled along a path can never be less than the magnitude of the net displacement between its endpoints, and it is strictly greater whenever the path is not a straight line. Since the cabman's route was circuitous ($\displaystyle 23$ km of driving to cover a $\displaystyle 10$ km straight-line gap), the two quantities differ — the only case in which they would come out equal is a trip made entirely along a straight line in one direction.Answer: (a) Average speed ≈ $\displaystyle 49$ km/h; (b) Magnitude of average velocity ≈ $\displaystyle 21$ km/h; the two are not equal because the actual path ($\displaystyle 23$ km) is longer than the straight-line displacement ($\displaystyle 10$ km).

A ball thrown at the "textbook" angle for maximum range, $\displaystyle 45$°, would rise more than $\displaystyle 40$ m — well above this $\displaystyle 25$ m ceiling — so $\displaystyle 45$° is not available here. The ball must be launched at a shallower angle, one whose highest point just grazes the ceiling.For a projectile launched with speed \(\displaystyle u \) at angle \(\displaystyle \theta \) above the horizontal (no air resistance), the two standard results of projectile motion are:Maximum height reached: \[H = \frac{u^2 \sin^2\theta}{2g} \]Horizontal range: \[R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g} \]Here \(\displaystyle u = 40 \text{ m s}^{-1} \), the ceiling limits the height to \(\displaystyle H = 25 \text{ m} \), and \(\displaystyle g = 9.8 \text{ m s}^{-2} \).Why $\displaystyle 45$° fails. At the angle that ordinarily gives the greatest range, \(\displaystyle \theta = 45^\circ \), the ball would rise to \[H_{45^\circ} = \frac{u^2}{4g} = \frac{(40)^2}{4 \times 9.8} = \frac{1600}{39.2} = 40.8 \text{ m}, \] far above the $\displaystyle 25$ m ceiling. A $\displaystyle 45$° throw hits the ceiling before the trajectory can develop into its full range.Finding the largest usable angle. As \(\displaystyle \theta \) increases from \(\displaystyle 0^\circ \) to \(\displaystyle 45^\circ \), \(\displaystyle \sin 2\theta \) keeps increasing, so in this range a bigger launch angle always means a longer range. That means the best strategy — among all angles that keep the ball under the $\displaystyle 25$ m ceiling — is to use the largest one allowed: the angle whose peak height is exactly $\displaystyle 25$ m. (If that peak angle turns out to be below $\displaystyle 45$°, as it will here, we know the range was still climbing when the ceiling stopped it, so grazing the ceiling truly is optimal — not an angle beyond $\displaystyle 45$°, where more height buys less extra range.)Set the maximum-height formula equal to the ceiling height and solve for \(\displaystyle \theta \): \[\frac{u^2 \sin^2\theta}{2g} = H \] \[\sin^2\theta = \frac{2gH}{u^2} = \frac{2 \times 9.8 \times 25}{(40)^2} = \frac{490}{1600} = 0.30625 \] \[\sin\theta = 0.5534, \qquad \cos\theta = \sqrt{1 - 0.30625} = \sqrt{0.69375} = 0.8329 \]This angle works out to about \(\displaystyle 33.6^\circ \), which is indeed below \(\displaystyle 45^\circ \) — confirming the range formula was still increasing when the ceiling cut the trajectory off, so this is the throw that goes farthest.Computing the range. Substitute \(\displaystyle \sin\theta \) and \(\displaystyle \cos\theta \) into the range formula: \[R = \frac{2u^2 \sin\theta \cos\theta}{g} = \frac{2 \times (40 \text{ m s}^{-1})^2 \times 0.5534 \times 0.8329}{9.8 \text{ m s}^{-2}} \] \[R = \frac{2 \times 1600 \times 0.4609 \text{ m}}{9.8} = \frac{1475.0 \text{ m}}{9.8} = 150.5 \text{ m} \]The given data — $\displaystyle 25$ m, $\displaystyle 40$ m s\(\displaystyle ^{-1}\), and \(\displaystyle g = 9.8 \) m s\(\displaystyle ^{-2}\) — each carry two significant figures, so the final distance is rounded, once, to two significant figures: \[R \approx 150 \text{ m} \]Answer: The ball can go at most about $\displaystyle 150$ m horizontally (thrown at roughly $\displaystyle 33.6$° to the horizontal) without hitting the $\displaystyle 25$ m ceiling.

The cricketer's real limit is how fast he can throw the ball, not the angle he throws it at — find that speed from the $\displaystyle 100$ m throw, then point the same speed straight up.For a ball launched with speed \(\displaystyle u\) at angle \(\displaystyle \theta\) above the horizontal, split the velocity into a horizontal part \(\displaystyle u\cos\theta\) and a vertical part \(\displaystyle u\sin\theta\). Gravity (acceleration \(\displaystyle g\), downward) acts only on the vertical part, so the horizontal part stays constant throughout the flight.Step $\displaystyle 1$ — where the range formula comes from. By symmetry, the ball returns to the same height after the vertical velocity has reversed sign, which takes time \(\displaystyle T = \dfrac{2u\sin\theta}{g}\). The horizontal distance covered in this time is horizontal speed \(\displaystyle \times\) time: \[R = (u\cos\theta)\,T = \frac{u^{2}(2\sin\theta\cos\theta)}{g} = \frac{u^{2}\sin 2\theta}{g} \] For a given throwing speed \(\displaystyle u\), \(\displaystyle R\) is largest when \(\displaystyle \sin 2\theta = 1\), i.e. at \(\displaystyle \theta = 45^{\circ}\). So the "maximum horizontal distance" the cricketer can achieve is \[R_{\text{max}} = \frac{u^{2}}{g} \] and this is the throw the question is describing: \(\displaystyle R_{\text{max}} = 100\ \text{m}\). This fixes the ball's launch speed: \[u^{2} = R_{\text{max}}\,g = (100\ \text{m})\,g \] This speed \(\displaystyle u\) is set by the cricketer's arm — it is the same no matter what angle he throws at.Step $\displaystyle 2$ — where the maximum-height formula comes from. Using the equation of motion \(\displaystyle v^{2} = u_{y}^{2} - 2gh\) for the vertical component (taking upward as positive, initial vertical speed \(\displaystyle u_{y}=u\sin\theta\), final vertical speed \(\displaystyle v=0\) at the top of the flight), the greatest height reached in a throw at angle \(\displaystyle \theta\) is \[H = \frac{u^{2}\sin^{2}\theta}{2g} \] This is largest when \(\displaystyle \sin\theta = 1\), i.e. \(\displaystyle \theta = 90^{\circ}\) — the ball thrown straight up. So "how high above the ground can the cricketer throw the ball" means using the same maximum speed \(\displaystyle u\) found in Step $\displaystyle 1$, but this time aimed straight up: \[H_{\text{max}} = \frac{u^{2}}{2g} \] A common mistake here is to read the height off the same $\displaystyle 45$° trajectory that gave the maximum range — but a $\displaystyle 45$° throw only reaches \(\displaystyle \dfrac{u^{2}}{4g}\), which is not the ball's true maximum possible height. The maximum range and the maximum height are two different throws of the same ball at the same top speed, at two different angles ($\displaystyle 45$° and $\displaystyle 90$°).Step $\displaystyle 3$ — combine, and watch \(\displaystyle g\) cancel. \[H_{\text{max}} = \frac{u^{2}}{2g} = \frac{(100\ \text{m})\,g}{2g} = \frac{100\ \text{m}}{2} = 50\ \text{m} \] Both formulas share the same factor \(\displaystyle u^{2}/g\) fixed in Step $\displaystyle 1$, so \(\displaystyle g\) cancels exactly and its numerical value never needs to be looked up — the maximum height a projectile can be thrown is always exactly half its maximum range, for any thrower. The given distance, $\displaystyle 100$ m, is taken to three significant figures, so the height is stated to match:Answer: $\displaystyle 50.0$ m — thrown straight up with the same top speed that sends it $\displaystyle 100$ m at $\displaystyle 45$°, the ball rises to half that distance, $\displaystyle 50.0$ m, because the maximum-range and maximum-height formulas share the identical factor \(\displaystyle u^{2}/g\).

Constant speed on a circle means the only acceleration is centripetal — pointed at the center, with magnitude \(\displaystyle a = \omega^2 r \).Since the stone's speed is constant, there is no tangential acceleration (nothing is speeding it up or slowing it down along the path). The only acceleration comes from the continuous change in the direction of the velocity, and for uniform circular motion this centripetal acceleration has magnitude\[a = \frac{v^2}{r} = \omega^2 r \]where \(\displaystyle r \) is the radius of the circle and \(\displaystyle \omega \) is the angular speed.Step $\displaystyle 1$: Convert the radius to metres.\[r = 80\ \text{cm} = 0.80\ \text{m} \]Step $\displaystyle 2$: Find the frequency of revolution.The stone completes $\displaystyle 14$ revolutions in $\displaystyle 25$ s, so the frequency (revolutions per second) is\[\nu = \frac{\text{number of revolutions}}{\text{time}} = \frac{14}{25\ \text{s}} = 0.56\ \text{s}^{-1} \]Step $\displaystyle 3$: Find the angular speed.Each revolution sweeps out an angle of \(\displaystyle 2\pi \) rad, so\[\omega = 2\pi\nu = 2\pi \left(\frac{14}{25}\right)\ \text{rad s}^{-1} = \frac{28\pi}{25}\ \text{rad s}^{-1} \approx 3.52\ \text{rad s}^{-1} \]Step $\displaystyle 4$: Find the centripetal acceleration.\[a = \omega^2 r = \left(3.52\ \text{rad s}^{-1}\right)^2 \times \left(0.80\ \text{m}\right) \]\[a \approx 12.38\ \text{m s}^{-2} \times 0.80\ \text{m} \approx 9.90\ \text{m s}^{-2} \]Step $\displaystyle 5$: Apply significant figures.Every given quantity ($\displaystyle 80$ cm, $\displaystyle 14$ rev, $\displaystyle 25$ s) carries two significant figures, so the answer is rounded to two significant figures at the end, not mid-calculation:\[a \approx 9.9\ \text{m s}^{-2} \]Direction: this acceleration points along the radius, toward the centre of the circle at every instant — that is the definition of "centripetal." It is easy to mix this up with the outward "pull" a person on a merry-go-round feels, but that sensation is inertia resisting the turn, not the true acceleration; the actual acceleration of the stone (and the tension in the string producing it) is always directed inward, toward the centre.Answer: The magnitude of the acceleration is about \(\displaystyle 9.9\ \text{m s}^{-2} \), directed along the string toward the centre of the circle (centripetal).

Centripetal acceleration comes from the speed and the radius of the turn, not from anything about the aircraft itself — the formula is \(\displaystyle a_c = v^2/r \).The aircraft moves at constant speed along a circle of radius \(\displaystyle r \), so it has a centripetal acceleration directed horizontally, toward the centre of the loop: \[a_c = \frac{v^2}{r} \] where \(\displaystyle v \) is the speed and \(\displaystyle r \) is the radius of the circular path.Step $\displaystyle 1$: Convert the speed to SI units.Never substitute a speed in km/h directly into a formula built for metres and seconds — convert first. Using \(\displaystyle 1\ \text{km/h} = \dfrac{5}{18}\ \text{m/s} \): \[v = 900\ \text{km/h} = 900 \times \frac{5}{18}\ \text{m/s} = 250\ \text{m/s} \]Step $\displaystyle 2$: Write the radius in metres. \[r = 1.00\ \text{km} = 1.00 \times 10^{3}\ \text{m} = 1000\ \text{m} \]Step $\displaystyle 3$: Substitute into \(\displaystyle a_c = v^2/r \). \[a_c = \frac{(250\ \text{m/s})^2}{1000\ \text{m}} = \frac{6.25\times 10^{4}\ \text{m}^2/\text{s}^2}{1.00\times 10^{3}\ \text{m}} = 62.5\ \text{m/s}^2 \]Step $\displaystyle 4$: Compare with the acceleration due to gravity.Taking \(\displaystyle g = 9.8\ \text{m/s}^2 \), form the ratio: \[\frac{a_c}{g} = \frac{62.5\ \text{m/s}^2}{9.8\ \text{m/s}^2} = 6.3776\ldots \]The data (\(\displaystyle 1.00\ \text{km} \), and \(\displaystyle 900\ \text{km/h} \) read to the same three-figure precision) supports three significant figures, so round once, at the end: \[\frac{a_c}{g} \approx 6.38 \]This is not a small correction — the centripetal acceleration is more than six times \(\displaystyle g \), and it points horizontally toward the centre of the loop, quite unlike the vertical acceleration due to gravity. A pilot flying this loop feels a horizontal push far stronger than their own weight, which is exactly why tight, fast turns are physically demanding on both aircraft and pilot.Answer: \(\displaystyle a_c = 62.5\ \text{m/s}^2 \approx 6.38\, g \) (about $\displaystyle 6.4$ times the acceleration due to gravity), directed horizontally toward the centre of the loop.

Circular-motion acceleration splits into two independent pieces — a radial part from the changing direction of velocity, and a tangential part from the changing speed — and each statement below turns on whether the tangential piece is actually present.(a) False.For a particle moving on a circle of radius \(\displaystyle r \) with speed \(\displaystyle v(t) \), the acceleration has, in general, two components: \[\vec{a} = \vec{a}_{\text{radial}} + \vec{a}_{\text{tangential}} \] where the radial (centripetal) part has magnitude \[a_{\text{radial}} = \frac{v^2}{r} \] and always points toward the centre — this piece exists simply because the direction of \(\displaystyle \vec v \) is continuously changing, and it is present whether or not the speed is constant.The tangential part is \[a_{\text{tangential}} = \frac{dv}{dt} \] directed along the tangent to the path (forward if the particle is speeding up, backward if slowing down). This piece is present whenever the speed itself is changing with time, i.e. whenever \(\displaystyle dv/dt \neq 0 \) — for instance, a car accelerating around a curve, or a stone on a string being swung faster and faster.The net acceleration is purely radial (along the centre) only in the special case \(\displaystyle dv/dt = 0 \), i.e. uniform circular motion. The statement claims this holds "always" for circular motion, which is too strong — for non-uniform circular motion the acceleration also has a tangential component, so it is not directed along the radius. The mistake to watch for is treating "circular path" as the same thing as "uniform circular motion"; the radius-only rule needs the extra condition of constant speed.(b) True.Velocity is defined as the instantaneous rate of change of position: \[\vec v = \lim_{\Delta t \to 0} \frac{\Delta \vec r}{\Delta t} \] As \(\displaystyle \Delta t \to 0 \), the displacement \(\displaystyle \Delta \vec r \) is a chord joining two points on the path that get closer and closer together, and the direction of that chord approaches the direction of the tangent to the path at that point. This is a property of the definition of velocity itself, true for a particle on any trajectory — straight line, circle, or any curve — not a special feature of circular motion. So the statement holds without exception.(c) True.In uniform circular motion the speed \(\displaystyle v \) is constant; only the direction of \(\displaystyle \vec v \) changes, sweeping around the circle once every period \(\displaystyle T \). At every instant the instantaneous acceleration has magnitude \(\displaystyle v^2/r \) directed toward the centre, so it is never zero — but its direction keeps rotating along with the particle.The average acceleration over a time interval is \[\langle \vec a \rangle = \frac{\vec v_{\text{final}} - \vec v_{\text{initial}}}{\Delta t} \] Take the interval to be exactly one cycle, \(\displaystyle \Delta t = T \). After one full revolution the particle is back at the same point on the circle, moving with the same speed in the same direction, so \[\vec v(T) = \vec v(0) \] Substituting, \[\langle \vec a \rangle = \frac{\vec v(0) - \vec v(0)}{T} = \vec 0 \] The average acceleration vector over one complete cycle is the null vector, even though the instantaneous acceleration is nonzero throughout — the instantaneous centripetal vectors point in every direction around the circle in turn, and over a full revolution they cancel out exactly. This is a good place to be careful about the difference between "instantaneous" and "averaged over a cycle": the first is never zero here, the second always is.Answer: (a) False — the acceleration is purely radial only for uniform circular motion; if the speed changes there is also a tangential component \(\displaystyle dv/dt\) along the tangent. (b) True — velocity is tangent to the path at every point, for any trajectory. (c) True — after one full period the velocity vector returns to its starting value, so the average acceleration over that cycle is the null vector.

Velocity is the rate of change of position, and acceleration is the rate of change of velocity — both come from differentiating the position vector with respect to time, component by component.The position vector is given as \[\vec{r}(t) = 3.0t\,\hat{i} - 2.0t^2\,\hat{j} + 4.0\,\hat{k} \ \text{m} \] where \(\displaystyle t\) is in seconds. Notice the \(\displaystyle \hat{k}\) term carries no \(\displaystyle t\) at all — the particle's z-coordinate never changes, so despite the three unit vectors this motion actually happens in a single plane (parallel to the xy-plane, offset by $\displaystyle 4.0$ m along z).(a) Finding \(\displaystyle \vec{v}\) and \(\displaystyle \vec{a}\)Velocity is \(\displaystyle \vec{v} = \dfrac{d\vec{r}}{dt}\). Differentiate each component on its own, using \(\displaystyle \frac{d}{dt}(t) = 1\), \(\displaystyle \frac{d}{dt}(t^2) = 2t\), and noting a plain constant like \(\displaystyle 4.0\hat{k}\) differentiates to zero: \[\vec{v}(t) = \frac{d}{dt}(3.0t)\,\hat{i} - \frac{d}{dt}(2.0t^2)\,\hat{j} + \frac{d}{dt}(4.0)\,\hat{k} \] \[\vec{v}(t) = 3.0\,\hat{i} - 4.0t\,\hat{j} \ \text{m/s} \]Acceleration is \(\displaystyle \vec{a} = \dfrac{d\vec{v}}{dt}\); differentiate once more: \[\vec{a}(t) = \frac{d}{dt}(3.0)\,\hat{i} - \frac{d}{dt}(4.0t)\,\hat{j} \] \[\vec{a}(t) = -4.0\,\hat{j} \ \text{m/s}^2 \]This acceleration has no \(\displaystyle t\) in it — it is constant, always \(\displaystyle 4.0\) m/s² in the \(\displaystyle -\hat{j}\) direction. So the particle moves at constant velocity along x while being uniformly accelerated along y; combined, the path traced in the xy-plane is a parabola.(b) Magnitude and direction of velocity at \(\displaystyle t = 2.0\) sSubstitute \(\displaystyle t = 2.0\) s into \(\displaystyle \vec{v}(t)\): \[\vec{v}(2.0) = 3.0\,\hat{i} - 4.0(2.0)\,\hat{j} = 3.0\,\hat{i} - 8.0\,\hat{j} \ \text{m/s} \]The two components, \(\displaystyle v_x = 3.0\) m/s and \(\displaystyle v_y = -8.0\) m/s, are perpendicular, so the magnitude follows from Pythagoras' theorem: \[|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (-8.0)^2} = \sqrt{9.0 + 64} = \sqrt{73} \ \text{m/s} \]Every given coefficient ($\displaystyle 3.0$, $\displaystyle 2.0$, $\displaystyle 4.0$) and the value \(\displaystyle t = 2.0\) s carry $\displaystyle 2$ significant figures, so the square root is kept unrounded until the very last step: \[|\vec{v}| = 8.544\ldots \ \text{m/s} \approx 8.5 \ \text{m/s} \]For the direction, use the two components to find the angle \(\displaystyle \theta\) the velocity makes with the positive x-axis: \[\tan\theta = \left|\frac{v_y}{v_x}\right| = \frac{8.0}{3.0} = 2.667 \] \[\theta = \tan^{-1}(2.667) \approx 69^\circ \]\(\displaystyle v_x = 3.0\) m/s is positive and \(\displaystyle v_y = -8.0\) m/s is negative, so the vector points to the right and downward — into the fourth quadrant. The velocity is therefore directed \(\displaystyle 69^\circ\) below the positive x-axis (equivalently, \(\displaystyle 69^\circ\) clockwise from the x-axis, or \(\displaystyle 291^\circ\) counter-clockwise from it). Quoting only "$\displaystyle 8.5$ m/s" would be an incomplete answer here — velocity is a vector, and stopping at the magnitude throws away half of what was asked.Answer: \(\displaystyle \vec{v}(t) = 3.0\hat{i} - 4.0t\,\hat{j}\) m/s and \(\displaystyle \vec{a}(t) = -4.0\hat{j}\) m/s² (constant); at \(\displaystyle t = 2.0\) s, \(\displaystyle \vec{v} = 3.0\hat{i} - 8.0\hat{j}\) m/s, magnitude \(\displaystyle 8.5\) m/s, directed \(\displaystyle 69^\circ\) below the positive x-axis.

Split the motion into two independent $\displaystyle 1$-D problems — one along x, one along y — because the acceleration has components in both directions while the initial velocity has a component only along y.The particle starts at the origin with initial velocity \(\displaystyle \vec{v_0} = 10.0\,\hat{j} \) m/s (zero component along x) and constant acceleration \(\displaystyle \vec{a} = (8.0\,\hat{i} + 2.0\,\hat{j}) \) m/s². Using the equation of motion \(\displaystyle \vec{r}(t) = \vec{r_0} + \vec{v_0}t + \tfrac{1}{2}\vec{a}t^2 \) separately for each axis (with \(\displaystyle x_0 = y_0 = 0 \)):\[x(t) = 0 + \tfrac{1}{2}(8.0)t^2 = 4.0\,t^2 \quad \text{(m)} \] \[y(t) = 10.0\,t + \tfrac{1}{2}(2.0)t^2 = 10.0\,t + 1.0\,t^2 \quad \text{(m)} \]Note the asymmetry that trips people up: the x-motion here starts from rest and is driven only by the $\displaystyle 8.0$ m/s² acceleration, while the y-motion already has a $\displaystyle 10.0$ m/s head start and only a small $\displaystyle 2.0$ m/s² acceleration added to it. The two axes must be tracked with their own numbers — don't carry the y initial velocity into the x equation or vice versa.(a) Time at which x = $\displaystyle 16$ m, and the y-coordinate thenSet \(\displaystyle x(t) = 16 \) m:\[4.0\,t^2 = 16 \implies t^2 = 4.0 \implies t = 2.0 \text{ s} \](the negative root is rejected because the particle starts moving at \(\displaystyle t = 0 \) s and time cannot run backward here).Substitute \(\displaystyle t = 2.0 \) s into the y-equation:\[y(2.0) = 10.0(2.0) + 1.0(2.0)^2 = 20.0 + 4.0 = 24.0 \text{ m} \]Rounding to two significant figures (set by the $\displaystyle 8.0$ and $\displaystyle 2.0$ m/s² data): \(\displaystyle t = 2.0 \) s and \(\displaystyle y = 24 \) m.(b) Speed of the particle at that timeSpeed is the magnitude of the velocity vector, not the vector itself — no direction is needed for this part, only the number. Get the two velocity components first, using \(\displaystyle v(t) = v_0 + at \) on each axis:\[v_x(2.0) = 0 + (8.0)(2.0) = 16 \text{ m/s} \] \[v_y(2.0) = 10.0 + (2.0)(2.0) = 10.0 + 4.0 = 14.0 \text{ m/s} \]The speed is then\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{(16)^2 + (14.0)^2} = \sqrt{256 + 196} = \sqrt{452} \approx 21.26 \text{ m/s} \]The least precise input feeding this result (\(\displaystyle v_x = 16 \) m/s, from the $\displaystyle 8.0$ m/s² acceleration) carries two significant figures, so the final speed is rounded to two significant figures at the very end: \(\displaystyle v \approx 21 \) m/s.Answer: (a) \(\displaystyle t = 2.0 \) s, at which the y-coordinate is \(\displaystyle 24 \) m. (b) The speed at that time is \(\displaystyle \approx 21 \) m/s.

Two unit vectors at right angles behave like two sides of a unit square — their sum and difference are that square's diagonals, and those diagonals are themselves perpendicular to each other.Take \(\displaystyle \hat{i} \) and \(\displaystyle \hat{j} \) as the unit vectors along the x- and y-axes, so in component form \(\displaystyle \hat{i} = (1,0) \) and \(\displaystyle \hat{j} = (0,1) \).Magnitude and direction of \(\displaystyle \hat{i}+\hat{j} \)Adding components, \[\hat{i}+\hat{j} = (1,1) \]By the Pythagorean theorem, the magnitude of a vector \(\displaystyle (x,y)\) is \(\displaystyle \sqrt{x^2+y^2} \), so \[|\hat{i}+\hat{j}| = \sqrt{1^2+1^2} = \sqrt{2} \]Its direction is the angle \(\displaystyle \theta \) it makes with the positive x-axis, found from \(\displaystyle \tan\theta = y/x \): \[\tan\theta = \frac{1}{1} = 1 \implies \theta = 45^\circ \]So \(\displaystyle \hat{i}+\hat{j} \) has magnitude \(\displaystyle \sqrt{2} \) and points at \(\displaystyle 45^\circ \) to the x-axis. It exactly bisects the right angle between \(\displaystyle \hat{i} \) and \(\displaystyle \hat{j} \) — which is why, drawn graphically, it is the diagonal of the unit square those two vectors form.Magnitude and direction of \(\displaystyle \hat{i}-\hat{j} \)\[\hat{i}-\hat{j} = (1,-1) \] \[|\hat{i}-\hat{j}| = \sqrt{1^2+(-1)^2} = \sqrt{2} \] \[\tan\theta = \frac{-1}{1} = -1 \implies \theta = -45^\circ \]So \(\displaystyle \hat{i}-\hat{j} \) also has magnitude \(\displaystyle \sqrt{2} \), but it points \(\displaystyle 45^\circ \) below the x-axis (into the fourth quadrant) rather than above it — the other diagonal of the same unit square.Check the two diagonals against each other: \[(\hat{i}+\hat{j})\cdot(\hat{i}-\hat{j}) = (1)(1) + (1)(-1) = 0 \] They are perpendicular. That is what makes them usable as a second, tilted pair of axes, which is exactly what the next part needs.Components of \(\displaystyle \mathbf{A} = 2\hat{i}+3\hat{j} \) along these two directionsThe component of a vector along a given direction is its dot product with the unit vector in that direction — not with \(\displaystyle \hat{i}+\hat{j} \) or \(\displaystyle \hat{i}-\hat{j} \) themselves, since each of those has length \(\displaystyle \sqrt{2} \), not 1. This is the step people skip. The unit vectors are \[\hat{n}_1 = \frac{\hat{i}+\hat{j}}{\sqrt{2}}, \qquad \hat{n}_2 = \frac{\hat{i}-\hat{j}}{\sqrt{2}} \]Component along \(\displaystyle \hat{i}+\hat{j} \): \[A_1 = \mathbf{A}\cdot\hat{n}_1 = \frac{(2\hat{i}+3\hat{j})\cdot(\hat{i}+\hat{j})}{\sqrt{2}} = \frac{2(1)+3(1)}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx 3.5 \]Component along \(\displaystyle \hat{i}-\hat{j} \): \[A_2 = \mathbf{A}\cdot\hat{n}_2 = \frac{(2\hat{i}+3\hat{j})\cdot(\hat{i}-\hat{j})}{\sqrt{2}} = \frac{2(1)+3(-1)}{\sqrt{2}} = \frac{-1}{\sqrt{2}} \approx -0.71 \]The negative sign on \(\displaystyle A_2 \) is not an error — it says \(\displaystyle \mathbf{A} \) leans strongly toward the \(\displaystyle \hat{i}+\hat{j} \) diagonal and slightly away from the \(\displaystyle \hat{i}-\hat{j} \) diagonal, rather than toward it.As a check, rebuilding \(\displaystyle \mathbf{A} \) from these two perpendicular components should return the original vector: \[A_1\hat{n}_1 + A_2\hat{n}_2 = \frac{5}{\sqrt{2}}\cdot\frac{\hat{i}+\hat{j}}{\sqrt{2}} + \left(\frac{-1}{\sqrt{2}}\right)\cdot\frac{\hat{i}-\hat{j}}{\sqrt{2}} = \frac{5}{2}(\hat{i}+\hat{j}) - \frac{1}{2}(\hat{i}-\hat{j}) = 2\hat{i}+3\hat{j} \] which matches \(\displaystyle \mathbf{A} \), confirming both components.Answer: \(\displaystyle \hat{i}+\hat{j} \) has magnitude \(\displaystyle \sqrt{2} \), directed at \(\displaystyle 45^\circ \) to the x-axis; \(\displaystyle \hat{i}-\hat{j} \) has magnitude \(\displaystyle \sqrt{2} \), directed at \(\displaystyle 45^\circ \) below the x-axis. Along \(\displaystyle \hat{i}+\hat{j} \), \(\displaystyle \mathbf{A}=2\hat{i}+3\hat{j} \) has component \(\displaystyle 5/\sqrt{2} \approx 3.5 \); along \(\displaystyle \hat{i}-\hat{j} \), its component is \(\displaystyle -1/\sqrt{2} \approx -0.71 \).

Two of these are definitions that hold no matter how the object moves; the other three secretly assume constant acceleration, which "arbitrary motion" does not give you.The phrase "arbitrary motion in space" means the acceleration can change with time in any way — speeding up, slowing down, turning, reversing. Only relations built directly from the definitions of average velocity and average acceleration survive that. Relations built by integrating an acceleration only work if you know what the acceleration is doing in between, and for arbitrary motion you don't.(a) \(\displaystyle v_{average} = \dfrac{1}{2}\left(v(t_1) + v(t_2)\right)\) — FALSEThis says the average velocity over an interval equals the simple mean of the velocities at the two endpoints. That is only true when velocity changes linearly with time, i.e., when the acceleration \(\displaystyle a\) is constant. For arbitrary motion the velocity–time graph can bulge, dip, or reverse between \(\displaystyle t_1\) and \(\displaystyle t_2\), so the endpoint average tells you nothing reliable about the true average velocity over the interval.(b) \(\displaystyle v_{average} = \dfrac{r(t_2) - r(t_1)}{t_2 - t_1}\) — TRUEThis is the definition of average velocity: total displacement divided by total time taken. Displacement \(\displaystyle r(t_2) - r(t_1)\) is a directly measurable quantity — the straight-line vector from the starting point to the ending point — regardless of how twisted the actual path was or how the speed varied along it. Because this is a definition, not a result derived from assuming constant acceleration, it holds for any motion whatsoever.(c) \(\displaystyle v(t) = v(0) + at\) — FALSEThis is the standard kinematic equation for velocity under constant acceleration \(\displaystyle a\). It comes from integrating \(\displaystyle a\) (treated as a fixed number) with respect to time. If the acceleration itself changes during the interval — which "arbitrary motion" explicitly allows — then the correct relation involves an integral of the actual acceleration function, \(\displaystyle v(t) = v(0) + \int_0^t a(t')\,dt'\), and the simple product \(\displaystyle at\) does not describe it.(d) \(\displaystyle r(t) = r(0) + v(0)t + \dfrac{1}{2}at^2\) — FALSESame reasoning as (c): this is the position equation obtained by integrating \(\displaystyle v(t) = v(0) + at\) once more, and it is valid only when \(\displaystyle a\) is constant throughout the motion. For arbitrary, time-varying acceleration, position must instead be found from \(\displaystyle r(t) = r(0) + \int_0^t v(t')\,dt'\), which reduces to the quadratic form only in the special case of uniform acceleration.(e) \(\displaystyle a_{average} = \dfrac{v(t_2) - v(t_1)}{t_2 - t_1}\) — TRUEThis is the definition of average acceleration: the change in velocity divided by the time taken for that change. Just like (b), it makes no assumption about how the acceleration behaved in between — it simply compares the velocity you started with to the velocity you ended with, over the known time interval. This holds for any arbitrary motion.A short aside to keep the two groups straight: (b) and (e) are definitions of average quantities — they only ever need the values at the two endpoints, so they are true by construction. (a), (c), and (d) are equations of uniformly accelerated motion — they were derived by assuming \(\displaystyle a\) stays constant, so they break down the moment the acceleration is allowed to vary, which is exactly what "arbitrary motion" permits.Answer: (b) and (e) are true for any arbitrary motion; (a), (c), and (d) are true only for motion with constant acceleration, not for arbitrary motion.

The one true test of a scalar is that its value does not depend on how you choose to orient your axes — everything else in this list sounds like a property of scalars but is not.(a) A scalar quantity is conserved in a process — False. Being conserved is a property of a specific physical law applied to a specific process, not a defining property of "being a scalar." Kinetic energy, for instance, is a perfectly good scalar quantity, but it is not conserved in an inelastic collision — some of it is converted to heat and sound. Total energy is conserved (a scalar that happens to be conserved), while total mechanical energy is not conserved once friction acts. So conservation depends on which quantity and which process you are looking at, not on the quantity being scalar.(b) A scalar quantity can never take negative values — False. Many scalars run negative routinely. Temperature measured on the Celsius or Fahrenheit scale can be below zero (for example \(\displaystyle -10\,^\circ\text{C} \)). Electric potential and potential energy can be negative depending on the choice of reference. Work done, \(\displaystyle W = \vec{F}\cdot\vec{d} \), is negative whenever the force has a component opposite to the displacement — pushing a block that friction is dragging backward does negative work. A scalar is just a single number with a magnitude and a sign convention; nothing forces that number to be non-negative. (This is different from magnitude of a vector, like speed, which by definition cannot be negative — do not confuse "scalar" with "magnitude.")(c) A scalar quantity must be dimensionless — False. Scalar only means the quantity is described completely by a single number (with a unit) and has no direction — it says nothing about dimensions. Mass \(\displaystyle [M] \), time \(\displaystyle [T] \), speed \(\displaystyle [LT^{-1}] \), work and energy \(\displaystyle [ML^2T^{-2}] \), and temperature \(\displaystyle [K] \) are all scalars, and all of them carry dimensions. Only a special subset of scalars — ratios such as the coefficient of friction, refractive index, or strain — happen to be dimensionless. Dimensionlessness is a coincidence for those particular quantities, not a requirement of being scalar.(d) A scalar quantity does not vary from one point to another in space — False. A scalar can be defined at every point of space and take a different value at each one; this is called a scalar field. Atmospheric temperature is a scalar (a single number, no direction) at every point in a room, yet it is warmer near a heater and cooler near a window — it clearly varies from point to point. Gravitational potential and electric potential are further examples of scalar fields that change with position. Being scalar restricts how many numbers describe the quantity at one point (namely, one), not whether that number can change as you move to another point.(e) A scalar quantity has the same value for observers with different orientations of axes — True. This is the actual defining property. When two observers set up their coordinate systems with the same origin but rotated relative to each other, a vector's components (like \(\displaystyle v_x, v_y, v_z \)) change from one frame to the other, because those components depend on the axes chosen. A scalar, by contrast, is not built from components along any axis at all — it is a single number tied to the physical situation itself. Mass, temperature, speed, and energy have one value regardless of which way you point your \(\displaystyle x \)-, \(\displaystyle y \)-, and \(\displaystyle z \)-axes, because rotating your axes changes nothing about the physics being described, only your bookkeeping of directions. This invariance under rotation of axes is precisely why physicists count something as a scalar.Answer: (a) False (b) False (c) False (d) False (e) True — the defining property of a scalar is that its value is unchanged under rotation of the coordinate axes, not conservation, non-negativity, dimensionlessness, or spatial uniformity.

The trick is to turn the $\displaystyle 30$° angle into two right triangles by dropping a perpendicular from the ground observer onto the aircraft's flight path.The aircraft flies level, so its two positions, call them \(\displaystyle P\) (at the first sighting) and \(\displaystyle Q\) ($\displaystyle 10.0$ s later), lie on the same horizontal line, at height \(\displaystyle h = 3400\ \text{m}\) above the observer's location \(\displaystyle O\). Drop a perpendicular from \(\displaystyle O\) onto that horizontal line, meeting it at \(\displaystyle M\) — this is the point directly overhead the observer, so \(\displaystyle OM = h = 3400\ \text{m}\) exactly, and \(\displaystyle OM\) is vertical while \(\displaystyle PQ\) is horizontal, so they meet at a right angle.The aside that matters here: the diagram behind this problem places the observer directly below the midpoint of the path swept out between the two sightings — the plane is as far past overhead at the second sighting as it was short of overhead at the first. That symmetry is what lets the perpendicular \(\displaystyle OM\) also bisect the angle at \(\displaystyle O\), splitting the given \(\displaystyle 30^\circ\) into two equal \(\displaystyle 15^\circ\) angles: \[\angle POM = \angle QOM = \frac{30^\circ}{2} = 15^\circ \]Now \(\displaystyle OMP\) and \(\displaystyle OMQ\) are right triangles, each with the ground-to-aircraft leg \(\displaystyle OM = h\) adjacent to the known angle, and \(\displaystyle PM\), \(\displaystyle QM\) the (horizontal) legs opposite it. Using the tangent ratio, \(\displaystyle \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\), in triangle \(\displaystyle OMP\): \[\tan 15^\circ = \frac{PM}{OM} \quad\Rightarrow\quad PM = h\tan 15^\circ \] and by the same reasoning in triangle \(\displaystyle OMQ\), \(\displaystyle QM = h\tan 15^\circ\). The distance the aircraft actually covered is \[PQ = PM + QM = 2h\tan 15^\circ \]\(\displaystyle \tan 15^\circ\) has the exact value \(\displaystyle \tan(45^\circ-30^\circ) = 2-\sqrt3 = 0.26795\) (kept to $\displaystyle 5$ figures here so the rounding happens only at the very end, not partway through).Substituting \(\displaystyle h = 3400\ \text{m}\): \[PQ = 2(3400\ \text{m})(0.26795) = 1822.1\ \text{m} \]Speed is distance covered divided by the time taken, \(\displaystyle v = \dfrac{PQ}{t}\), with \(\displaystyle t = 10.0\ \text{s}\) the given interval between the two sightings: \[v = \frac{1822.1\ \text{m}}{10.0\ \text{s}} = 182.2\ \text{m/s} \]The time was given to three significant figures ($\displaystyle 10.0$ s), so the answer is rounded to three significant figures too: \(\displaystyle v \approx 182\ \text{m/s}\). The question asks only for speed, not velocity, so no direction needs to be reported — just this magnitude, which works out to roughly $\displaystyle 656$ km/h, a plausible cruising speed for an aircraft.Answer: \(\displaystyle v \approx 182\ \text{m/s}\) (≈ $\displaystyle 656$ km/h)