Kinematics solutions

Model Earth and Moon as spherically symmetric. The gravitational-force magnitude is F = Gm_Em_M/r².

Why: Spherical symmetry lets us treat each body's gravity as coming from its center. Newton's inverse-square law uses the center-to-center separation.

Convert r = 384 400 km to 3.84400 × 10⁸ m.

Why: The stated G uses metres, so convert kilometres to metres before squaring the separation.

Substitute F = (6.673 × 10⁻¹¹)(5.97 × 10²⁴)(7.36 × 10²²)/(3.84400 × 10⁸)².

Why: Put both masses and the converted separation into the same gravitational-force model. Squaring r gives the required inverse-square dependence.

The result is 1.98430 × 10²⁰ N, directed from the Moon toward Earth’s center.

Why: The calculation gives the force's magnitude. Gravity is attractive, so the force on the Moon points toward Earth's center.

Resolve v₀ into v₀ₓ = 10 cos 55° = 5.73576 m/s and v₀ᵧ = 10 sin 55° = 8.19152 m/s.

Why: The launch angle starts at horizontal, so cosine gives the constant horizontal component and sine gives the initial upward component.

At the peak vᵧ = 0, so 0 = v₀ᵧ² − 2gΔy; Δy_max = v₀ᵧ²/(2g) = 3.42352 m.

Why: At maximum height only the vertical velocity is zero. Use the constant vertical acceleration −g in the equation without time to find the rise.

At the catch height, 0 = v₀ᵧt − ½gt². Discard t = 0 and solve t = 2v₀ᵧ/g = 1.67174 s.

Why: Returning to launch height makes the net vertical displacement zero. The t = 0 root is launch itself; the other root is the later catch.

Integrate acceleration and impose vₓ(0) = 0: vₓ(t) = t⁴/7 + (5/2)t² + 12t, in m/s.

Why: Velocity is the time integral of acceleration. Starting from rest fixes the integration constant at zero.

For t ≥ 0 every term is nonnegative, so vₓ never becomes negative and distance equals displacement.

Why: Distance equals displacement only when motion never reverses. All velocity terms are nonnegative over 0–4 s, so that condition holds here.

Integrate velocity from 0 to 4 s: d = [t⁵/35 + (5/6)t³ + 6t²]₀⁴.

Why: Position change is the area under the velocity curve. Integrate the given polynomial over the actual 0–4 s interval.

Evaluate: d = 1024/35 + 160/3 + 96 = 178.59048 m.

Why: Evaluate the antiderivative at 4 s and subtract its value at 0. The three terms add the separate contributions to travelled distance.

At the highest point the vertical velocity is zero, while horizontal velocity remains constant.

Why: Gravity changes the vertical component, but with no air resistance the horizontal component stays unchanged. At the peak, that component is the whole speed.

Thus v_peak = v₀ cos θ, so cos θ = 60/80 = 0.75.

Why: Because the launch angle is measured above horizontal, the unchanging horizontal speed is v₀ cos θ. Its ratio to v₀ isolates cos θ.

For the usual forward launch with 0° < θ < 90°, θ = arccos(0.75).

Why: The stated upward forward launch restricts θ to the first quadrant. Choose the arccos value in that range.

For equal launch and landing height, vertical motion gives t = 2v₀ sin θ/g.

Why: The equal-height condition lets the vertical motion return to zero displacement. Its nonzero time root is 2v₀ sin θ/g.

Multiply by horizontal velocity v₀ cos θ: R = v₀² sin(2θ)/g.

Why: Horizontal speed stays constant in this ideal model. Multiplying it by flight time and using 2 sin θ cos θ = sin 2θ gives the range.

With launch velocity fixed, lower g gives a longer flight and a larger range.

Why: Keeping speed and angle fixed holds the numerator constant. Lower g takes longer to undo the upward motion, so the projectile travels farther sideways.

Mars has lower gravitational acceleration than Earth and Jupiter. With the stated representative values, R_Mars/R_Earth = 9.8/3.71 = 2.642 and R_Mars/R_Jupiter = 24.79/3.71 = 6.682.

Why: Range is inversely proportional to g, so each range ratio uses the other planet's g over Mars's g. These numbers use the stated representative accelerations.

Distance is ∫|vₓ|dt. This graph is never negative, so use the ordinary area under vₓ(t).

Why: Distance counts speed, the absolute value of velocity. Since this graph never goes below zero, its signed area already equals distance.

From 0 to 2 s the triangular hump has area ½(2 s)(1 m/s) = 1 m.

Why: The first hump is a triangle with a 2 s base and 1 m/s height. Its area has units (s)(m/s) = m.

From 2 to 3 s the triangle has area ½(1)(2) = 1 m.

Why: During 2–3 s the velocity rises from zero, making a second triangle with a 1 s base and 2 m/s height.

From 3 to 5 s the rectangle has area (2)(2) = 4 m.

Why: The 3–5 s velocity is constant, so this graph segment is a rectangle: speed times elapsed time.

From 5 to 6 s the triangle has area ½(1)(2) = 1 m. Sum 1 + 1 + 4 + 1.

Why: The final slowing segment is another positive triangle. Add all four nonoverlapping graph areas to get the full distance.

Convert separation to r = 2.43 m. Gravity requires m₁m₂ = Fr²/G = 2477.70418 kg².

Why: Convert the separation to SI units and rearrange F = Gm₁m₂/r². The measured force requires a particular product of the two masses.

The stated mass sum gives m₂ = 70.5 − m₁. Thus m₁² − 70.5m₁ + 2477.70418 = 0.

Why: The stated total mass leaves only one independent unknown. Substituting m₂ = 70.5 − m₁ turns the product constraint into a quadratic.

The discriminant is 70.5² − 4(2477.70418) = −4940.56672 kg², so this quadratic has no real roots.

Why: A quadratic needs a nonnegative discriminant for real roots. The negative value shows the stated force and total mass cannot both hold.

Equivalently, a fixed positive sum of 70.5 kg permits product at most (70.5/2)² = 1242.5625 kg², attained by equal masses.

Why: For a fixed positive sum, unequal masses reduce the product. Equal masses give the largest possible product, providing a second consistency check.

The largest possible force at this separation is G(35.25 kg)²/(2.43 m)² = 1.40419 × 10⁻⁸ N, smaller than the stated force.

Why: Insert that largest allowed product into the gravity law. Even this maximum force falls below the stated force, confirming the contradiction.

For a low-mass planet orbiting the Sun, T_planet² = a³ in Earth years and AU. Thus T_planet = √(13³) = 46.87217 years = 1.47928558 × 10⁹ s.

Why: For a negligible-mass body orbiting the Sun, AU and Earth years reduce Kepler's relation to T² = a³. Convert the planet year to seconds for the probe calculation.

The stated 5000 orbits imply T_probe = T_planet/5000 = 295 857.12 s and speed 2π(79 500 m)/T_probe = 1.68836 m/s.

Why: Five thousand equal probe periods must fit in one planet year. Dividing gives the claimed period, and circumference over period gives its implied speed.

For a freely orbiting circular probe, GM/r² = v²/r, so v = √(GM/r) = √[(6.673 × 10⁻¹¹)(3.4 × 10¹⁸)/79 500] = 53.42155 m/s.

Why: A circular gravitational orbit needs gravity to supply v²/r. Equating those accelerations tests the claimed speed against the given mass and radius.

That circular speed gives period 2πr/v = 9350.41 s, or approximately 158 205 probe orbits per planet year.

Why: Use the gravity-required speed to calculate one actual circular period. Dividing the planet year by that period predicts the compatible orbit count.

The two speeds differ by a factor of about 31.64, so the orbit count, mass and radius cannot all describe this circular gravitational orbit.

Why: The same circular orbit cannot have both speeds. Their large ratio shows that the stated count, mass and radius are mutually inconsistent.

Take launch horizontal position x₀ = 0, ground y = 0, and up as +y. Components are v₀ₓ = 28 cos 35° = 22.93626 m/s and v₀ᵧ = 28 sin 35° = 16.06014 m/s.

Why: Setting ground to y = 0 keeps the 50 m launch height explicit. Project the launch velocity using its angle above horizontal and choose upward as positive.

At 5 s, x = v₀ₓt = 114.68129 m and vᵧ = v₀ᵧ − gt = −32.93986 m/s.

Why: Horizontal velocity stays constant, so x = v₀ₓt. Vertical velocity changes by −gt over the same 5 s because gravity acts downward.

The velocity vector is (22.93626, −32.93986) m/s. Its speed is √(22.93626² + 32.93986²) = 40.13859 m/s.

Why: Keep the horizontal and vertical velocity components together to state direction. Their perpendicular squared components add to give speed.

The velocity angle is atan2(−32.93986,22.93626) = −55.15023°, or 55.15023° below horizontal.

Why: atan2 uses both component signs, locating the rightward, downward velocity below horizontal instead of in the wrong quadrant.

Ground impact solves 0 = 50 + 16.06014t − 4.9t². The positive root is t_hit = 5.22901 s; remaining time is 5.22901 − 5 = 0.22901 s.

Why: Ground impact is y = 0, so solve the height equation for its future positive root. Subtract the already elapsed 5 s to get time remaining.

The height at 5 s is 50 + 16.06014(5) − 4.9(25) = 7.80070 m, confirming these values occur before impact.

Why: A positive height at 5 s checks that the projectile has not already struck the ground. The earlier velocity calculations still describe free flight.

Let P = (x,y) lie on x²/a² + y²/b² = 1. Its distances to (−c,0) and (+c,0) are d₋ = √[(x+c)²+y²] and d₊ = √[(x−c)²+y²].

Why: Distance from a point to a focus follows Pythagoras. Writing both distances explicitly gives the two expressions whose sum we need to prove.

From the ellipse equation, y² = b²(1 − x²/a²), and c² = a² − b².

Why: Use the ellipse equation to eliminate y². The focus relation c² = a² − b² then lets every distance depend on a, c and x.

Substitute into the first squared distance: (x+c)²+y² = a² + 2cx + c²x²/a² = (a+cx/a)². Similarly, (x−c)²+y² = (a−cx/a)².

Why: Expand each squared distance and substitute the ellipse relations. Recognizing a perfect square makes the square roots manageable.

Since |x| ≤ a and 0 ≤ c < a, both a+cx/a and a−cx/a are positive. Therefore the square roots give d₋ = a+cx/a and d₊ = a−cx/a.

Why: A square root is an absolute value, so check the signs before dropping it. The ellipse's |x| bound and c < a make both expressions positive.

Add the distances: d₋+d₊ = 2a. The major axis runs from (−a,0) to (a,0), so its length is also 2a.

Why: The cx/a terms cancel when the distances add. The remaining 2a equals the separation of the two major-axis endpoints.

Distance adds path lengths: 30 + 18 = 48 m; displacement uses signs: 30 − 18 = +12 m.

Why: Travelled lengths add for distance; opposite directions subtract for displacement. East-positive makes the west leg negative only in the displacement sum.

Elapsed time includes the wait: 10 + 5 + 6 = 21 s.

Why: Average motion covers the entire interval, including the 5 s with no movement. Leaving out the wait would overstate both averages.

Average speed = 48/21; average velocity = 12/21.

Why: Divide distance by the full time for average speed, and signed displacement by that same time for average velocity.

The runner’s position is x_r = 4t; for t ≥ 6 s the cyclist’s position is x_c = 10(t − 6).

Why: The runner has moved for t seconds, while the cyclist has moved for only t − 6. Both positions use the marker as the same origin.

Set positions equal: 4t = 10t − 60, so t = 10 s.

Why: Catching means occupying the same position at the same time. Equate the two position functions to find that time.

The cyclist has ridden for 4 s, and the meeting position is 4(10) = 40 m.

Why: The reported t starts with the runner, so subtract the 6 s delay for the cyclist's riding time. Either position function gives the meeting location.

Choose east positive and set final velocity to zero: 0 = 18 − 1.5t, so t = 12 s.

Why: Stopping means v = 0. East-positive gives a negative braking acceleration, so v₀ + at reaches zero after a positive time.

Use v² = v₀² + 2aΔx: 0 = 18² − 3Δx, so Δx = 108 m.

Why: With constant acceleration, the squared-velocity relation finds displacement directly without using time. The negative a correctly removes the initial v².

The required 108 m exceeds the 100 m available by 8 m.

Why: Compare the required stopping distance with the actual room available. The extra 8 m is why the tram cannot meet the 100 m limit.

After acceleration v = 1.2(10) = 12 m/s; the first stage covers ½(1.2)(10²) = 60 m.

Why: Starting from rest makes v = at and distance = ½at² during the constant-acceleration stage.

The cruise stage covers 12(20) = 240 m.

Why: The train keeps its reached 12 m/s speed during cruise, so multiply that speed by the cruise duration.

During uniform braking the average speed is (12 + 0)/2 = 6 m/s, giving distance 6(8) = 48 m.

Why: Uniform braking makes velocity vary linearly. The average of its endpoint speeds gives the mean speed for this stage.

Sum distance 348 m and time 38 s; average speed = 348/38.

Why: The three stages happen consecutively, so add their distances and times. Whole-trip average speed uses those totals.

Differentiate: vₓ(t) = 6 − 2t m/s and aₓ(t) = −2 m/s².

Why: Velocity is the slope of position versus time, and acceleration is the slope of velocity. Differentiating the polynomial gives both functions.

The velocity changes sign at t = 3 s. Positions are x(0) = 2 m, x(3) = 11 m and x(5) = 7 m.

Why: A turn requires velocity to pass through zero and change sign. Evaluate position there and at both interval endpoints to split the path correctly.

Displacement = 7 − 2 = +5 m. Distance = (11 − 2) + (11 − 7) = 13 m.

Why: Displacement subtracts the initial position from the final one. Distance adds the forward trip to the turning point and the return trip separately.

Velocity change is acceleration times interval: v(4) = 3 + 2(4) = 11 m/s.

Why: For the first constant-acceleration interval, velocity changes by aΔt. Add that change to the initial 3 m/s.

For the second interval v(10) = 11 − 1(6) = 5 m/s.

Why: The second interval lasts 6 s and starts at 11 m/s. Its negative acceleration reduces velocity by 6 m/s.

Displacements are 3(4) + ½(2)(4²) = 28 m and 11(6) − ½(1)(6²) = 48 m.

Why: Calculate each interval's displacement using its own initial velocity and constant acceleration. The second stage starts at 11 m/s, not the original 3 m/s.

Velocity remains positive, so displacement and distance are both 76 m.

Why: Even after slowing, velocity is still positive. The sled never reverses, so its two positive displacements add and also equal distance.

The first rectangle contributes +12 m. In the linear stage a = (−2 − 4)/3 = −2 m/s², so velocity is zero at t = 5 s.

Why: Area under velocity gives displacement. The first stage is a rectangle; the later straight-line slope gives acceleration and locates the zero-velocity crossing.

The positive triangle from 3 to 5 s has area ½(2)(4) = 4 m; the negative triangle from 5 to 6 s has signed area −½(1)(2) = −1 m.

Why: Split the changing-velocity stage at t = 5 s. The area before the crossing is positive, while the below-axis area after it is negative.

Displacement = 12 + 4 − 1 = 15 m. Distance = 12 + 4 + 1 = 17 m.

Why: Signed areas cancel for displacement. Distance counts both forward and backward travel positively, so use the magnitude of the negative triangle.

Write x(t) = v₀t + ½at² since x₀ = 0. The first two observations give 1 = v₀ + ½a and 4 = 2v₀ + 2a.

Why: The constant-acceleration assumption permits x = x₀ + v₀t + ½at². Use two measured positions to form equations for the two unknowns.

Subtract twice the first equation from the second: 2 = a, in m/s². Then v₀ = 0.

Why: Eliminate v₀ by subtracting twice the first equation. Once a is known, substitute it back to recover the initial velocity.

Check the remaining observation: x(3) = ½(2)(3²) = 9 m. The data agree.

Why: The third observation was not used to solve for v₀ and a. Matching it checks that one constant-acceleration model fits all the data.

If total time is T, downward travel is s(t) = ½gt². The last-second distance is s(T) − s(T − 1) = ½g(2T − 1).

Why: Measure downward travel from release. The last-second distance is the difference between total travel at T and travel one second earlier, not the full drop height.

Set 24.5 = 4.9(2T − 1), so T = 3 s.

Why: Set that interval distance equal to 24.5 m and solve for T. The expansion of T² − (T − 1)² leaves a linear equation.

Drop height is ½(9.8)(3²) = 44.1 m.

Why: Now use the full fall time in the rest-start displacement formula. This gives the whole drop height rather than just its last second.

Height relative to release is y = 14t − 4.9t². Solve 4.9t² − 14t + 5 = 0.

Why: Use the release point as the height origin and upward as positive. Setting y to 5 m produces the quadratic for both visits to that height.

The roots are t = (14 ± √98)/9.8 = 0.41842 s and 2.43872 s.

Why: The quadratic has two positive roots because the ball passes 5 m once on the way up and once on the way down.

At each root vᵧ = 14 − 9.8t: +9.89949 m/s on ascent and −9.89949 m/s on descent.

Why: Velocity changes by −gt from its upward launch value. Its opposite signs at the roots distinguish ascent from descent.

Vertical motion has initial vertical velocity zero: 45 = ½gt², so t = √(90/9.8) = 3.03046 s.

Why: A horizontal launch has no initial vertical velocity. The 45 m drop and gravity therefore determine flight time independently of horizontal speed.

Horizontal range is x = 12t = 36.36549 m.

Why: The package keeps its 12 m/s horizontal velocity in the ideal model. Multiplying by the same fall time gives the range.

At impact vᵧ = −gt = −29.69848 m/s and vₓ = 12 m/s. Speed = √(12² + 29.69848²).

Why: Gravity supplies the downward impact component while the horizontal component remains 12 m/s. Pythagoras combines them into the impact speed.

Eliminate time from x = v₀ cos θ·t and y = v₀ sin θ·t − ½gt² to obtain y = x tan θ − gx²/(2v₀² cos²θ).

Why: Both components share one flight time. Solve horizontal motion for t and substitute it into vertical motion to describe the trajectory at the target's x.

Rearrange: v₀² = gx²/[2cos²θ(x tan θ − y)]. The denominator is positive, so the target is reachable at this angle.

Why: Isolate v₀² in the trajectory equation. A positive x tan θ − y is needed because gravity can only bend the shot below its gravity-free launch line.

Substitute x = 20 m, y = 4 m and θ = 40°: v₀ = 16.16467 m/s.

Why: Use the target coordinates and fixed angle in that expression, then take the positive square root because the requested quantity is speed.

Equal-height range is R = v₀² sin 2θ/g. Since sin 60° = sin 120°, both ranges are 22.62270 m.

Why: For equal landing height, range depends on sin 2θ. The complementary launch angles give the same value, balancing horizontal speed against flight time.

Maximum height is H = v₀² sin²θ/(2g). The 30° shot reaches 3.26531 m; the 60° shot reaches 9.79592 m, three times as high.

Why: Peak height depends on the square of the initial vertical speed. The 60° shot has three times the squared vertical component of the 30° shot.

Flight time is 2v₀ sin θ/g: 1.63265 s and 2.82784 s.

Why: The time to return to launch height depends on the initial vertical component. The steeper launch stays airborne longer even though its range is the same.

The ball inherits the trolley’s ground-frame horizontal velocity: vₓ = 3 m/s. Its vertical launch velocity is 9.8 m/s.

Why: Relative velocities add: a straight-up throw in the trolley's frame still carries the trolley's 3 m/s horizontal motion in the ground frame.

Flight time is 2v₀ᵧ/g = 2 s. Ground-frame horizontal displacement is 3(2) = 6 m east.

Why: Vertical motion gives the equal-height return time. Multiply that time by the unchanged ground-frame horizontal velocity to find the ball's ground displacement.

The trolley also travels 6 m during this time, so relative horizontal displacement is zero.

Why: The ball and trolley keep the same horizontal velocity. Subtracting their equal ground displacements gives no sideways displacement relative to the trolley.

The initial line of sight has length √(12² + 5²) = 13 m, so launch components are v₀ₓ = 26(12/13) = 24 m/s and v₀ᵧ = 26(5/13) = 10 m/s.

Why: The line-of-sight displacement is a 5–12–13 triangle. Divide its components by 13 to get a direction, then scale that direction by the 26 m/s launch speed.

The ball reaches the target’s unchanged horizontal coordinate at t = 12/24 = 0.5 s.

Why: The target has no horizontal motion, while the ball's horizontal speed stays 24 m/s. Covering the 12 m horizontal gap fixes the meeting time.

Both objects lose the same vertical height ½gt² = 1.225 m relative to their gravity-free positions. Thus both are at y = 5 − 1.225 = 3.775 m.

Why: Both objects share acceleration −g, so gravity lowers both by the same amount in the same time. Their gravity-free paths meet, so their falling paths meet too.

The target would hit the ground at √(2·5/g) = 1.01015 s, later than the meeting.

Why: The target's ground-hit time comes from dropping 5 m from rest. Comparing it with 0.5 s confirms the meeting occurs while both are above ground.

The mutual force magnitude is F = G(2)(6)/(0.50²) = 3.20304 × 10⁻⁹ N, equal on both spheres and directed toward the other sphere.

Why: Apply the inverse-square law to the center separation. The two forces have equal magnitudes by Newton's third law and point toward each other.

Acceleration magnitudes are F/2 = 1.60152 × 10⁻⁹ m/s² and F/6 = 5.33840 × 10⁻¹⁰ m/s².

Why: Each acceleration is its own force divided by its own mass. Equal interaction forces therefore give the lighter sphere the larger acceleration.

Doubling separation divides force by four. With masses fixed, each acceleration is also divided by four.

Why: The separation is squared in the denominator, so doubling it cuts force to one quarter. Fixed masses make each acceleration scale by the same factor.

Gravity provides centripetal acceleration: GM/r² = v²/r, so v = √(GM/r).

Why: For a circular orbit, the gravitational acceleration must equal the inward v²/r. Cancelling one factor of r gives the circular-speed expression.

Substitute the data: v = √[(6.673 × 10⁻¹¹)(8.0 × 10²²)/(2.0 × 10⁶)] = 1633.77 m/s.

Why: Insert the planet mass and center-to-center orbital radius in SI units. The result is in m/s before converting it to km/s.

Period is circumference divided by speed: T = 2πr/v = 7691.65 s = 2.13657 h.

Why: At constant circular speed, one orbit covers circumference 2πr. Divide by speed to get seconds, then by 3600 to express hours.

Since g(r) = GM/r², the ratio g(r)/g(R) = (R/r)².

Why: Compare the gravity law at r and at the surface radius R. The common GM cancels, leaving a pure ratio of squared distances.

Set (R/r)² = 1/4, giving r = 2R. Height above the surface is h = r − R = R.

Why: Quarter-strength gravity needs twice the center distance. Subtract the original surface radius to convert that distance into height above the surface.

The acceleration magnitude there is 9.8/4 = 2.45 m/s².

Why: The requested acceleration is one quarter of the stated surface value. Its direction stays inward because Earth's gravity is attractive.

Focus distance c = √(a² − b²) = √7 = 2.64575 AU. Eccentricity e = c/a = 0.661438.

Why: An ellipse's focus offset follows c² = a² − b². Dividing c by a gives the dimensionless eccentricity rather than another distance.

With the Sun at a focus, perihelion r_p = a − c = 1.35425 AU and aphelion r_a = a + c = 6.64575 AU.

Why: The Sun sits at a focus, not the ellipse's center. The near and far major-axis endpoints are therefore a − c and a + c from the Sun.

Solar Kepler units give T² = a³, so T = √(4³) = 8 years.

Why: For a negligible-mass solar orbiter in AU and Earth years, Kepler's third law is T² = a³. The period depends on the semimajor axis.

The semimajor axis is a = (0.80 + 5.20)/2 = 3.00 AU. Solar Kepler units give T² = a³, hence T = √27 = 5.19615 Earth years.

Why: Perihelion plus aphelion spans the major axis, so halve their sum for a. Use that a, not either endpoint radius, in the solar period relation.

At perihelion and aphelion, velocity is perpendicular to the radius, so equal areal speed gives r_pv_p = r_av_a.

Why: At the closest and farthest points, velocity is tangent and perpendicular to radius. Equal areas swept in equal times then requires the products rv to match.

The speed ratio is v_p/v_a = r_a/r_p = 5.20/0.80 = 6.50.

Why: Rearrange the equal-area relation to compare speeds. The shorter perihelion radius requires the greater speed by the inverse radius ratio.