A radius vector of a point A relative to the origin varies with time t as r = ati − bt2j, where a and b are positive constants, and i and j are the unit vectors of the x and y axes. Find: (a) the equation of the point’s trajectory y(x); plot this function; (b) the time dependence of the velocity v and acceleration w vectors, as well as of the moduli of these quantities; (c) the time dependence of the angle α between the vectors w and v; (d) the mean velocity vector averaged over the first t seconds of motion, and the modulus of this vector.
The path is the parabola y = −(b/a²)x². The velocity is v = a i − 2bt j with |v| = √(a² + 4b²t²), and the acceleration is w = −2b j with |w| = 2b. The angle between them has tan α = a/(2bt), and the mean velocity is ⟨v⟩ = a i − bt j with |⟨v⟩| = √(a² + b²t²). With the numbers on screen: y = − x² and w = m/s². At t = s: α = ° and |⟨v⟩| = m/s.
Problem statement
Problem 1.24. A radius vector of a point A relative to the origin varies with time t as r = ati − bt2j, where a and b are positive constants, and i and j are the unit vectors of the x and y axes. Find: (a) the equation of the point’s trajectory y(x); plot this function; (b) the time dependence of the velocity v and acceleration w vectors, as well as of the moduli of these quantities; (c) the time dependence of the angle α between the vectors w and v; (d) the mean velocity vector averaged over the first t seconds of motion, and the modulus of this vector.
Interpretation. The vector r points from the origin to the point A. Its x part grows steadily (at) and its y part is negative and grows as t², so the point drifts to the right and falls. Part (a) asks for the path as a function y(x), with t removed. Part (b) asks for the velocity and acceleration as vectors and for their sizes. Part (c) asks for the angle between w and v. Part (d) asks for the mean velocity over the first t seconds, which is a vector, and for its size.
We assume a = 3 m/s and b = 0.5 m/s², because the problem gives no numbers. The units follow from r = at i − bt² j: a is a speed and b is an acceleration over two. Both are sliders. The graphs run from t = 0 to 6 s.
Symbols: r is the position vector, v the velocity, w the acceleration, α the angle between w and v, ⟨v⟩ the mean velocity, i and j the unit vectors along x and y.
The big idea in plain words
Treat the two directions separately. Sideways, x = at, so the speed a never changes. Downward, y = −bt², so the point falls faster and faster, like a stone thrown sideways. Put the two together and you get a thrown-stone path: a parabola. The acceleration is the same at every moment and points straight down. The velocity starts out sideways and tilts more and more toward down, so the angle between w and v starts at 90° and shrinks.
The solution in each frame
Only the ground frame is needed. The problem gives r in one frame and asks for everything in that frame, so there is no second frame.
The solution step by step
Step 1, the path (part a). The coordinates are x = at and y = −bt². From the first, t = x/a. Put this into the second: y = −b(x/a)² = −(b/a²)x². This is a parabola that opens downward, with its top at the origin. Here y = − x². The plot is the bright and dashed curve on the canvas.
Step 2, velocity and acceleration (part b). Compare x = at and y = −bt² with s = ut + ½At². For x there is no t² term, so ux = a and Ax = 0. For y there is no t term, so uy = 0, and ½Ay = −b gives Ay = −2b. Then v = u + At = a i − 2bt j. Its size is |v| = √(a² + 4b²t²). The acceleration is w = A = −2b j. It does not depend on t, and |w| = 2b = m/s².
Step 3, the angle (part c). The angle between two vectors comes from the dot product: w·v = wv cos α. Here w·v = (−2b)(−2bt) = 4b²t, so cos α = 4b²t/(2b√(a² + 4b²t²)) = 2bt/√(a² + 4b²t²). Then sin α = a/√(a² + 4b²t²), so tan α = a/(2bt) and α = arctan(a/(2bt)). At t = 0 the angle is 90°. For large t it goes toward 0°. It is 45° at t = a/(2b) = s.
Step 4, the mean velocity (part d). The mean velocity over the first t seconds is the displacement divided by the time. The displacement is r(t) − r(0) = at i − bt² j, so ⟨v⟩ = a i − bt j. Its size is |⟨v⟩| = √(a² + b²t²). Compare with the speed √(a² + 4b²t²): the mean velocity is smaller, because the point was slower earlier. The y part of ⟨v⟩ is exactly half of the final y velocity, as for constant acceleration.
Step 5, the cases. The problem has one situation, so no cases are needed. One thing to notice is that the angle never reaches 0° or goes below 0°. The velocity always keeps a sideways part a, so v never points exactly along w.
A separate calculation does not use the formulas for v, w, α or ⟨v⟩. It takes the position r(t) and finds the velocity and acceleration by small differences (steps of 0.001 s). It gets the angle from the dot product of those two numbers. It finds the mean velocity by adding up v dt in 200000 pieces and dividing by t. It does this at t = 1 s, 3 s and 6 s. The largest relative difference from the formulas is . At the moment shown on the canvas, ⟨v⟩ = (, ) m/s.
Five methods
1. Force method
No mass or force is given, so this method does not apply. The problem is about motion only (kinematics).
2. Energy method
Energy does not help here, since no mass is given.
3. Graph method
The first graph shows the two velocity parts and the speed. The sideways part is a flat line at a. The downward part is a straight line through zero with slope −2b. The speed is the green curve that starts at a and bends upward. The second graph shows α falling from 90°. The third graph shows |⟨v⟩|, which always stays below |v|.
4. Algebraic method (suvat)
Write x and y next to s = ut + ½At² and read off u = (a, 0) and A = (0, −2b). Then v = u + At and w = A. This is step 2. It is the fastest route to parts (b) and (c), with no calculus.
5. Check method
Units: a is in m/s, so at is in metres. b is in m/s², so bt² is in metres, as it should be. Limits: for b going to 0 the point moves in a straight line along x, and there is no angle between w and v, because w is zero. For a going to 0 the point falls straight down, v is along w, and α = 0. The formula tan α = a/(2bt) gives 0 for a = 0, which agrees. At t = 0, ⟨v⟩ = a i, the same as v, as it must be.
Fastest: the algebraic method. Most reliable: the algebraic method together with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Constant a | Yes | The width of the path, v, α, ⟨v⟩ and every curve |
| Constant b | Yes | The depth of the path, v, w, α, ⟨v⟩ and every curve |
| Moment shown t | No | The point, the arrows, the angle arc, the dots on the graphs and the live readings. The path and w do not depend on it |
| Given-answer mode | Sets t | The page finds the time at which α has a chosen value, and warns if that time is outside 0.05 to 6 s |
| Play and time bar | No | Play sweeps t from 0 to 6 s, holds 2 s at α = 45° (if it happens within 6 s) and 1.5 s at the end, then repeats. The bar sets the time |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Sideways speed | a | m/s |
| Constant in y = −bt² | b | m/s² |
| Path coefficient | b/a² | 1/m |
| Acceleration size | 2b | m/s² |
| Time when α = 45° | a/(2b) | s |
Exam traps
- Giving |v| as the sum a + 2bt. The two parts are at right angles, so the size is the square root of the sum of squares.
- Forgetting the minus sign in y = −bt². The point falls, so vy and w both point down (negative j).
- Writing tan α = 2bt/a. That is the angle between v and the x axis. The angle between w and v is measured from the vertical, so it is the other way up: tan α = a/(2bt).
- Using the final velocity as the mean velocity. The mean velocity is displacement divided by time: a i − bt j, with half the downward part of v.
- Averaging the speed instead of the velocity. The mean velocity is a vector, and its size √(a² + b²t²) is not the same as the average speed along the curved path.