University Physics I · course UPI
University Physics I
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
SI Quantities, Units & Prefixes: infer an input with uncertainty
An SI measurement is a number multiplied by a unit, where every coherent derived unit is a product of powers of seven base units with no extra numerical factor, and a prefix multiplies a unit by an exact power of ten. Use the supplied topic model Y_out = log(n)+k*p, where Y_out is LOG₁₀ OF VALUE IN SI (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. n is Number n in front of the unit, dimensionless. Keep k (Prefix 10ᵏ (milli −3, centi −2, kilo 3)) = -2; p (Power p on the unit (2 area, 3 volume)) = 2 fixed. For this exercise use only 1.90312 ≤ n ≤ 8.59688 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -3.2228406, with a bounded reading uncertainty ±0.023577. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000669375, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
LOG₁₀ OF VALUE IN SI against n
Predictions from the supplied SI Quantities, Units & Prefixes model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): LOG₁₀ OF VALUE IN SI
Horizontal axis (x): n: Number n in front of the unit
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Number n in front of the unit | LOG₁₀ OF VALUE IN SI |
|---|---|
| 1.903125 | -3.7205326856870307 |
| 3.5765625 | -3.446534181309344 |
| 5.25 | -3.279840696594043 |
| 6.9234375 | -3.1596782239522754 |
| 8.596875 | -3.0656593879302223 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(n) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.90312) = -3.72053 and Y_out(8.59688) = -3.06566. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 1.90312 + (-3.22284 − -3.72053)(8.59688 − 1.90312)/(-3.06566 − -3.72053) = 6.99026. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives n = 5.98631; substituting back gives Y_out = -3.2228406.
- The interpolation error in the input is |n_line − n| = 1.00394. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -3.2464176 and -3.1992636. Sort the roots even if the response decreases: 5.66999 ≤ n ≤ 6.32028.
- The lower and upper input deviations are 0.316321 and 0.333968. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer n ≈ 5.98631; compatible interval [5.66999, 6.32028]. Endpoint interpolation gives 6.99026.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Position, Displacement & Distance: infer an input with uncertainty
Position x is a signed coordinate measured from a chosen origin along a chosen positive direction; displacement Δx = x(t₂) − x(t₁) is its change over an interval; distance d is the total length of path covered, and is never negative. Use the supplied topic model Y_out = s*(2+6*t2-t2^2-x0), where Y_out is POSITION x(t₂) (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t2 is End of interval t₂, measured in s. Keep x0 (Origin at track mark x₀) = 5 m; s (Positive direction: +1 right, −1 left) = 1 fixed. For this exercise use only 1.03125 s ≤ t2 ≤ 2.86875 s on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 5.281108 m, with a bounded reading uncertainty ±0.10525 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00018375 s, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
POSITION x(t₂) against t2
Predictions from the supplied Position, Displacement & Distance model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): POSITION x(t₂) / m
Horizontal axis (x): t2: End of interval t₂ / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t2: End of interval t₂ / s | POSITION x(t₂) / m |
|---|---|
| 1.03125 | 2.1240234375 |
| 1.490625 | 3.7217871093750006 |
| 1.95 | 4.897499999999999 |
| 2.409375 | 5.651162109374997 |
| 2.86875 | 5.982773437499999 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(t2) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.03125) = 2.12402 m and Y_out(2.86875) = 5.98277 m. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t2_line = 1.03125 + (5.28111 − 2.12402)(2.86875 − 1.03125)/(5.98277 − 2.12402) = 2.53462 s. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives t2 = 2.15213 s; substituting back gives Y_out = 5.281108 m.
- The interpolation error in the input is |t2_line − t2| = 0.382499 s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.175858 m and 5.386358 m. Sort the roots even if the response decreases: 2.09218 s ≤ t2 ≤ 2.21665 s.
- The lower and upper input deviations are 0.0599477 s and 0.0645219 s. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t2 ≈ 2.15213 s; compatible interval [2.09218, 2.21665] s. Endpoint interpolation gives 2.53462 s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Vector Position, Velocity & Acceleration: infer an input with uncertainty
The position vector r⃗(t) locates a particle from a chosen origin at each instant; its first time derivative is the velocity v⃗, tangent to the path, and its second is the acceleration a⃗, both taken component by component. Use the supplied topic model Y_out = abs(deg(atan(amp*(cos(t+dt)-cos(t))/dt)-atan(-amp*sin(t)))), where Y_out is CHORD OFF TANGENT (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; deg converts radians to degrees; unmarked trigonometric functions use radians. amp is Amplitude A, measured in m. Keep t (Time t) = 2 s; dt (Averaging interval Δt) = 1.5 s fixed. For this exercise use only 0.39125 m ≤ amp ≤ 1.71425 m on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 24.884823 °, with a bounded reading uncertainty ±0.25451 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown amp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001323 m, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CHORD OFF TANGENT against amp
Predictions from the supplied Vector Position, Velocity & Acceleration model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CHORD OFF TANGENT / °
Horizontal axis (x): amp: Amplitude A / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| amp: Amplitude A / m | CHORD OFF TANGENT / ° |
|---|---|
| 0.39125 | 11.855027951701976 |
| 0.722 | 19.22530748313177 |
| 1.05275 | 23.688419511201182 |
| 1.3835 | 25.882350720955355 |
| 1.7142499999999998 | 26.581557928058707 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(amp) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.39125) = 11.855 ° and Y_out(1.71425) = 26.5816 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives amp_line = 0.39125 + (24.8848 − 11.855)(1.71425 − 0.39125)/(26.5816 − 11.855) = 1.56182 m. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives amp = 1.19828 m; substituting back gives Y_out = 24.884823 °.
- The interpolation error in the input is |amp_line − amp| = 0.363539 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 24.630313 ° and 25.139333 °. Sort the roots even if the response decreases: 1.16286 m ≤ amp ≤ 1.2372 m.
- The lower and upper input deviations are 0.0354162 m and 0.0389158 m. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer amp ≈ 1.19828 m; compatible interval [1.16286, 1.2372] m. Endpoint interpolation gives 1.56182 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Force Models & Inertial Frames: infer an input with uncertainty
A force model represents each interaction between two objects by a force vector whose size comes from a measured force law, and it applies in an inertial frame — one in which an object with no net interaction keeps a constant velocity. Use the supplied topic model Y_out = 2*u^2*f*(1-f)/g, where Y_out is HEIGHT ABOVE DECK, BOTH (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². f is Moment in the flight t/T, dimensionless. Keep vc (Flatcar speed V) = 3 m s⁻¹; u (Throw speed u, relative to the flatcar) = 5 m s⁻¹ fixed. For this exercise use only 0.52625 ≤ f ≤ 0.89375 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.95457349 m, with a bounded reading uncertainty ±0.047419 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00003675, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
HEIGHT ABOVE DECK, BOTH against f
Predictions from the supplied Force Models & Inertial Frames model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): HEIGHT ABOVE DECK, BOTH / m
Horizontal axis (x): f: Moment in the flight t/T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Moment in the flight t/T | HEIGHT ABOVE DECK, BOTH / m |
|---|---|
| 0.52625 | 1.2706979485219163 |
| 0.618125 | 1.2030911537971456 |
| 0.71 | 1.0494393476044854 |
| 0.8018749999999999 | 0.809742529943935 |
| 0.8937499999999999 | 0.4840007008154946 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(f) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.52625) = 1.2707 m and Y_out(0.89375) = 0.484001 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 0.52625 + (0.954573 − 1.2707)(0.89375 − 0.52625)/(0.484001 − 1.2707) = 0.673925. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives f = 0.750425; substituting back gives Y_out = 0.95457349 m.
- The interpolation error in the input is |f_line − f| = 0.0764997. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.90715449 m and 1.0019925 m. Sort the roots even if the response decreases: 0.731104 ≤ f ≤ 0.768359.
- The lower and upper input deviations are 0.019321 and 0.0179335. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer f ≈ 0.750425; compatible interval [0.731104, 0.768359]. Endpoint interpolation gives 0.673925.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.