University Physics I · unit UPI-02
1D Kinematics
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.
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.
Average and instantaneous velocity: infer an input with uncertainty
Average velocity is displacement divided by elapsed time, the slope of a secant on the x–t graph; instantaneous velocity is the limit of that ratio as the interval shrinks to zero, the slope of the tangent, dx/dt. Use the supplied topic model Y_out = 1.5*t^2-9*t+12, where Y_out is TANGENT SLOPE v(t) (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t is Instant t, measured in s. Keep d (Window Δt) = 1 s; c (Window: 0 forward, 1 centred) = 0 fixed. For this exercise use only 1.19312 s ≤ t ≤ 2.94137 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 = -0.67761736 m s⁻¹, with a bounded reading uncertainty ±0.12258 m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000174825 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.
TANGENT SLOPE v(t) against t
Predictions from the supplied Average and instantaneous velocity 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): TANGENT SLOPE v(t) / m s⁻¹
Horizontal axis (x): t: Instant t / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Instant t / s | TANGENT SLOPE v(t) / m s⁻¹ |
|---|---|
| 1.193125 | 3.3971958984374986 |
| 1.6301875 | 1.314579427734376 |
| 2.06725 | -0.19496615625000047 |
| 2.5043125 | -1.1314408535156275 |
| 2.941375 | -1.4948446640625015 |
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(t) = 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(1.19312) = 3.3972 m s⁻¹ and Y_out(2.94137) = -1.49484 m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 1.19312 + (-0.677617 − 3.3972)(2.94137 − 1.19312)/(-1.49484 − 3.3972) = 2.64933 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 t = 2.25956 s; substituting back gives Y_out = -0.67761736 m s⁻¹.
- The interpolation error in the input is |t_line − t| = 0.389768 s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.80019736 m s⁻¹ and -0.55503736 m s⁻¹. Sort the roots even if the response decreases: 2.20629 s ≤ t ≤ 2.31697 s.
- The lower and upper input deviations are 0.0532672 s and 0.0574088 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 t ≈ 2.25956 s; compatible interval [2.20629, 2.31697] s. Endpoint interpolation gives 2.64933 s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Average and instantaneous acceleration: infer an input with uncertainty
Acceleration is the rate of change of velocity: over an interval its average is a_avg = Δv/Δt, at an instant it is a = dv/dt, and both carry a sign along the declared positive axis. Use the supplied topic model Y_out = (10000*(v0 - 3*t0 + 0.5*t0^2)*(t0 - 3))/10000, where Y_out is va (+ SPEEDING UP) (before display rounding) (m² s⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t0 is Instant t₀, measured in s. Keep dt (Averaging interval Δt) = 1.5 s; v0 (Starting velocity v₀) = 2.5 m s⁻¹ fixed. For this exercise use only 0.425 s ≤ t0 ≤ 1.79 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 = 0.84000129 m² s⁻³, with a bounded reading uncertainty ±0.10413 m² s⁻³. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001365 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.
va (+ SPEEDING UP) (before display rounding) against t0
Predictions from the supplied Average and instantaneous 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): va (+ SPEEDING UP) (before display rounding) / m² s⁻³
Horizontal axis (x): t0: Instant t₀ / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t0: Instant t₀ / s | va (+ SPEEDING UP) (before display rounding) / m² s⁻³ |
|---|---|
| 0.425 | -3.386929687500001 |
| 0.76625 | -1.1053031279296874 |
| 1.1075 | 0.3959523984374998 |
| 1.44875 | 1.2360541787109367 |
| 1.79 | 1.5342195 |
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(t0) = 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(0.425) = -3.38693 m² s⁻³ and Y_out(1.79) = 1.53422 m² s⁻³. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t0_line = 0.425 + (0.840001 − -3.38693)(1.79 − 0.425)/(1.53422 − -3.38693) = 1.59744 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 t0 = 1.25765 s; substituting back gives Y_out = 0.84000129 m² s⁻³.
- The interpolation error in the input is |t0_line − t0| = 0.339792 s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.73587129 m² s⁻³ and 0.94413129 m² s⁻³. Sort the roots even if the response decreases: 1.21846 s ≤ t0 ≤ 1.30027 s.
- The lower and upper input deviations are 0.0391927 s and 0.0426204 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 t0 ≈ 1.25765 s; compatible interval [1.21846, 1.30027] s. Endpoint interpolation gives 1.59744 s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Position, Velocity & Acceleration Graphs: infer an input with uncertainty
Motion graphs plot position, velocity and acceleration against one time axis: reading down from x to v to a means taking a slope, and reading back up means adding a signed area to a known starting value. Use the supplied topic model Y_out = 2*(v0*min(tp,-v0/a)+0.5*a*min(tp,-v0/a)^2)-(v0*tp+0.5*a*tp^2), where Y_out is DISTANCE = UNSIGNED AREA (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. a is Constant acceleration a, measured in m/s². Keep v0 (Initial velocity v₀) = 8 m/s; tp (Probe time t) = 6 s fixed. For this exercise use only -1.86875 m/s² ≤ a ≤ -1.60625 m/s² 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 = 20.21227 m, with a bounded reading uncertainty ±0.049092 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00002625 m/s², 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.
DISTANCE = UNSIGNED AREA against a
Predictions from the supplied Position, Velocity & Acceleration Graphs 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): DISTANCE = UNSIGNED AREA / m
Horizontal axis (x): a: Constant acceleration a / m/s²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Constant acceleration a / m/s² | DISTANCE = UNSIGNED AREA / m |
|---|---|
| -1.86875 | 19.884991638795988 |
| -1.803125 | 19.950184142114388 |
| -1.7375 | 20.10953237410072 |
| -1.671875 | 20.374123831775698 |
| -1.6062500000000002 | 20.75685797665369 |
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(a) = 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(-1.86875) = 19.885 m and Y_out(-1.60625) = 20.7569 m. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = -1.86875 + (20.2123 − 19.885)(-1.60625 − -1.86875)/(20.7569 − 19.885) = -1.77021 m/s². 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 a = -1.70862 m/s²; substituting back gives Y_out = 20.21227 m.
- The interpolation error in the input is |a_line − a| = 0.0615887 m/s². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 20.163178 m and 20.261362 m. Sort the roots even if the response decreases: -1.7217 m/s² ≤ a ≤ -1.69659 m/s².
- The lower and upper input deviations are 0.0130707 m/s² and 0.0120386 m/s². 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 a ≈ -1.70862 m/s²; compatible interval [-1.7217, -1.69659] m/s². Endpoint interpolation gives -1.77021 m/s².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.