At the moment t = 0 a particle leaves the origin and moves in the positive direction of the x axis. Its velocity varies with time as v = v0(1 − t/τ), where v0 is the initial velocity vector whose modulus equals v0 = 10.0 cm/s; τ = 5.0 s. Find: (a) the x coordinate of the particle at the moments of time 6.0, 10, and 20 s; (b) the moments of time when the particle is at the distance 10.0 cm from the origin; (c) the distance s covered by the particle during the first 4.0 and 8.0 s; draw the approximate plot s(t).
(a) x = 24, 0 and -200 cm at 6, 10 and 20 s. (b) The particle is 10 cm from the origin at 5 − √15 = 1.13 s, 5 + √15 = 8.87 s and 5 + √35 = 10.9 s. (c) The path covered is 24 cm in 4 s and 34 cm in 8 s. With the numbers on screen: (b) is 1.127, 8.873 and 10.92 s for D = 10 cm, and (c) is 24 cm and 34 cm.
Problem statement
Problem 1.21. At the moment t = 0 a particle leaves the origin and moves in the positive direction of the x axis. Its velocity varies with time as v = v0(1 − t/τ), where v0 is the initial velocity vector whose modulus equals v0 = 10.0 cm/s; τ = 5.0 s. Find: (a) the x coordinate of the particle at the moments of time 6.0, 10, and 20 s; (b) the moments of time when the particle is at the distance 10.0 cm from the origin; (c) the distance s covered by the particle during the first 4.0 and 8.0 s; draw the approximate plot s(t).
Interpretation. The particle starts at the origin at t = 0 and moves along the x axis. The velocity v is the x component of the velocity, so it can become negative. The distance from the origin in (b) is measured along the axis, so the particle can be at x = +10 cm or at x = −10 cm. The distance s in (c) is the total path covered, which is not the same as the position x.
We assume v0 = 10 cm/s and τ = 5 s, as given. Both are sliders, so you can see how the answers change.
We assume D = 10 cm in part (b), as given. It is a slider. The times 6, 10 and 20 s in part (a) and 4 and 8 s in part (c) are fixed, as in the problem.
Symbols: x is the position, v the velocity along x, s the total path covered, τ the time when the velocity becomes zero.
The big idea in plain words
The velocity falls steadily from v0 to zero at t = τ, and then turns negative. So the particle goes out, stops at its farthest point, and comes back. The position is the area under the velocity graph, counted with sign. The path covered is the same area counted without sign, because walking back still adds to the distance walked. That one difference explains parts (a) and (c).
The solution in each frame
Only the ground frame is needed. The origin is fixed and the particle moves along one line, so there is no second frame in the problem.
The solution step by step
Step 1, the position. The velocity is v = dx/dt, so x = ∫v dt from 0 to t. With the particle at the origin at t = 0, x = ∫v0(1 − t/τ)dt = v0t(1 − t/2τ). With the numbers on screen, x = 10 t (1 − t/10). The farthest point is at t = τ, where v = 0: x = v0τ/2 = 25 cm. The particle is back at the origin at t = 2τ = 10 s.
Step 2, part (a). Put t = 6, 10 and 20 s into x = v0t(1 − t/2τ). For the given numbers, x(6) = 10 × 6 × (1 − 6/10) = 24 cm, x(10) = 10 × 10 × (1 − 10/10) = 0, and x(20) = 10 × 20 × (1 − 20/10) = −200 cm. On screen: 24, 0 and -200 cm. The negative sign at 20 s means the particle is on the negative side of the origin.
Step 3, part (b). The particle is at distance D from the origin when x = +D or x = −D. For x = +D, v0t − v0t²/2τ = D, which is t² − 2τt + 2τD/v0 = 0. So t = τ ± √(τ² − 2τD/v0). For the given numbers this is 5 ± √15 s, which is 1.13 s and 8.87 s: the particle passes x = +10 cm going out and again coming back. For x = −D the equation is t² − 2τt − 2τD/v0 = 0, and the root that is not negative is t = τ + √(τ² + 2τD/v0) = 5 + √35 = 10.9 s. The other root is negative, which is before the start, so it is dropped. On screen: 1.127, 8.873 and 10.92 s. If D is larger than v0τ/2, the square root has no real value and the particle never reaches +D.
Step 4, part (c). The path covered is the sum of |v| dt. Up to t = τ the velocity is positive, so s = x. So s(4 s) = x(4 s) = 10 × 4 × (1 − 4/10) = 24 cm. After t = τ the particle moves back, so the path adds the way back: s = xmax + (xmax − x) = v0τ/2 + v0(t − τ)²/2τ. At t = 8 s: s = 25 + 10 × 3²/10 = 34 cm. On screen: 24 cm and 34 cm.
Step 5, the plot of s(t). The third graph shows it. From 0 to τ the curve rises along x = v0t(1 − t/2τ), with a slope that falls to zero at t = τ. After that the slope grows again as the speed grows in the other direction, so the curve bends upward. It never goes down, because the path covered cannot decrease. The graph shows it up to 20 s, where s = 250 cm for the given numbers.
A separate calculation does not use the formulas. It adds up v dt in 200000 small steps over 20 s. It finds x(6) = 24, x(10) = -6.326 × 10⁻¹⁵ and x(20) = -200 cm, against 24, 0 and -200 cm from the formula. Adding up |v| dt gives s(4) = 24 and s(8) = 34 cm, against 24 and 34 cm. It then looks for the steps where x crosses +D and −D, and finds t = 1.127, 8.873 and 10.916 s, against 1.127, 8.873 and 10.92 s from the formulas. The largest relative difference is 8.4 × 10⁻¹¹.
Five methods
1. Force method
This problem gives the velocity, not forces, so the force method does not apply. The acceleration is dv/dt = −v0/τ, a constant, so the particle moves as if a constant force acts against its first motion.
2. Energy method
Energy does not help here, since no mass or force is given.
3. Graph method
Draw v against t. It is a straight line from v0 at t = 0 down to zero at t = τ and on to negative values. The position x is the signed area under the line. Up to τ the area is the triangle v0τ/2. After that the area below the axis is subtracted. The path s adds that area instead of subtracting it. This gives the answers with no integral.
4. Calculus method
x = ∫v dt and s = ∫|v| dt. Split the second integral at t = τ where v changes sign. This is steps 1 to 4.
5. Check method
Units: v0t is cm/s times s, which is cm. Limits: at t = 0 the position is zero. At t = 2τ the position is zero again, and the path is v0τ, twice the farthest distance. Symmetry: the positions at τ − d and τ + d are equal, which is why x(6) and x(4) are both 24 cm for τ = 5 s.
Fastest: the graph method. Most reliable: the calculus method with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Initial speed v0 | Yes | The positions, the farthest point, the times in (b), the distances in (c), all curves and the picture |
| Time constant τ | Yes | The turn-back moment, the return to the origin, all times, all curves and the picture |
| Distance D | Yes | The two dashed lines and the times in (b). Not (a) or (c) |
| Moment shown t | No | The particle, the arrows, the dots on the graphs and the live readings. Not the answers to (a), (b), (c) |
| Given-answer mode | Sets t | The page finds the time t for a chosen path length s, and warns if it is beyond 20 s |
| Play and time bar | No | Play runs the motion from 0 to 20 s, holds 2 s at the turning point and 1.2 s at the end, then repeats |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Initial speed | v0 | 10 cm/s |
| Time constant | τ | 5 s |
| Farthest point | v0τ/2 | 25 cm |
| Position at 6, 10, 20 s | x | 24, 0, -200 cm |
| Distance from the origin | D | 10 cm |
| Times for +D | t | 1.127 s, 8.873 s |
| Time for −D | t | 10.92 s |
| Path at 4 s and 8 s | s | 24 cm, 34 cm |
Exam traps
- Giving the position when the question asks for the path. At 20 s the position is −200 cm, but the path covered is 250 cm.
- Forgetting that the particle turns back at t = τ. After that the position falls but the path still grows.
- Using only the positive root in (b). The distance 10 cm can also mean x = −10 cm, which gives a third time.
- Keeping the negative root of the quadratic for x = −D. A negative time is before the start.
- Computing s(8 s) as x(8 s). That gives 16 cm, which is the position, not the path. The path is 34 cm.