The law is x = (35/π) sin πt, so a = 35/π cm and T = 2 s. In 2.80 s the point makes n = 5 quarter periods plus 0.30 s, and the distance is s = (35/π)(6 − cos 54°) ≈ 60 cm. For your sliders a = cm, n = , φ = degrees and s = cm.
Problem statement
At the moment t = 0 a particle starts moving along the x axis so that its velocity projection varies as vx = 35 cos πt cm/s, where t is in seconds. Find the distance that this particle covers during t = 2.80 s after the start.
We assume the particle starts at x = 0 at t = 0. This means the position is the integral of vx, and only the distance matters for the answer, so the start position does not change it.
We assume the motion goes on exactly as vx = v0 cos ωt for all time. This means a and ω stay fixed, with v0 = 35 cm/s and ω = π rad/s.
We assume "distance" means the total path length. This means the trip back counts as well as the trip out.
Symbols: v0 is the speed constant, ω the angular frequency, a = v0/ω the amplitude, T = 2π/ω the period, n the number of whole quarter periods in t, and φ the leftover phase.
The big idea in plain words
A velocity of the form v0 cos ωt belongs to a harmonic oscillation about x = 0. Integrating once gives the position law and shows the amplitude is v0/ω. Then the problem is the one we know: every quarter of a period covers exactly a, and only the leftover needs a formula. This time the particle starts at the middle, where the velocity is largest.
Interactive derivation
The solution in each frame
Lab frame
The axis is fixed to the lab, with x = 0 at the equilibrium position. The full solution is in the next section.
Point rest frame
This frame does not apply. The point keeps accelerating, so a frame fixed to it is not inertial, and it does not help to find the path length.
The solution step by step
First we find the motion law. The given velocity is vx = 35 cos πt, which has the form vx = aω cos ωt for x = a sin ωt. So we read off ω and a.
ω = π s⁻¹, a ω = 35 cm/s → a = 35/π cm ≈ 11.14 cm
T = 2π/ω = 2 s, T/4 = 0.5 s
Now we write the time in quarter periods. We need the whole number of quarters and what is left over.
t = 2.80 s = 5 × (T/4) + 0.30 s → n = 5, t0 = 0.30 s
φ = ω t0 = π × 0.30 = 0.3π = 54°
The particle starts at the middle, because x = a sin ωt is zero at t = 0. After an odd number of quarters it is at an extreme (n = 1 is the extreme x = a). So with n = 5 the leftover starts at an extreme, and it has moved from a down to a cos φ from there. Each full quarter covers exactly a.
s = n a + a(1 − cos φ) = a [n + 1 − cos φ]
s = (35/π)(5 + 1 − cos 54°) = 11.14 × (6 − 0.5878) = 11.14 × 5.412 = 60.3 cm ≈ 60 cm
When n is even the leftover starts at the middle and the formula changes to s = a[n + sin φ]. The page picks the right case by itself. For your sliders: a = cm, n = , so n a = cm, the leftover is cm, and s = cm.
Book check
Book answer: about 60 cm. The live value for the book setting is 60.3 cm, which matches to the book's rounding. The PDF has several typos in this solution: "x = a shoot" for x = a sin ωt, "(35/22) × 7" for 35/π with π taken as 22/7, and "interval 0 = 3 s", which should read 0.3 s. Its result 5 × (35/π) + (35/π)[1 − cos 0.3π] is the same as ours. A second formula (the area under the speed graph) gives the same distance, and a numeric integral of |vx| over the window agrees: cm against cm. A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
The law vx = v0 cos ωt gives wx = −v0 ω sin ωt = −ω² x, which is a spring-like force. So the motion is simple harmonic and the amplitude formula a = v0/ω holds. Force reasoning gives the type of motion, but the distance still needs the quarter-period count.
2. Energy method
At the middle the speed is the top speed v0, and at the ends it is zero. Energy gives (1/2) m v0² = (1/2) m ω² a², so a = v0/ω, the same result. It gives the amplitude quickly but not the distance for a given time.
3. Graph method
The distance is the area under the speed against time graph. Each quarter period has the same area a, so the distance graph climbs by a at each dotted line. The leftover is a part of one more hump.
4. Calculus method
This is the step-by-step route: s = ∫ |vx| dt with |vx| = 35 |cos πt|. Over each quarter the integral is a, and the leftover is a sin φ or a(1 − cos φ).
5. Check method
Units: a is a length, so s is a length. Limits: at t = T/4 the formula gives s = a, at t = T it gives 4a, as it must. For small t the distance is about v0 t, because the particle starts at its top speed. The distance never decreases, and it is never less than the displacement.
Fastest: the graph method. Most reliable: the calculus method, with the numeric check.
What each slider changes
Control
Changes the answer?
What it changes
Speed constant v0
Yes, proportional
The amplitude a = v0/ω, every distance, the track, the tape, the speed graph height
Angular frequency ω
Yes
a = v0/ω, T, the number of quarters n in t, the leftover phase, the distance and the motion speed
Time t
Yes
n, the leftover phase, the leftover distance, the total s and the tape
Screen speed
No
Only how fast the animation runs compared with real time
Target distance (given-answer mode)
Sets the answer
Solves t from s by bisection
Zoom, Auto-fit
No
Pixels per centimetre in the picture only
Parameters
Quantity
Symbol
Value used
Speed constant
v0
cm/s (book: 35)
Angular frequency
ω
rad/s (book: π)
Amplitude
a = v0/ω
cm
Period
T
s
Time
t
s (book: 2.80)
Whole quarter periods
n
Leftover phase
φ
degrees
Distance
s
cm
Exam traps
Integrating vx from 0 to t to get the distance. That gives the displacement, which here is 35/π × sin(2.8π) ≈ 6.5 cm, not 60 cm.
Using a = 35 cm instead of 35/π cm. The amplitude is the speed constant divided by ω.
Using the even-n leftover a sin φ for n = 5. Here n is odd, so the leftover starts at an extreme and is a(1 − cos φ).
Counting n with half periods. A quarter period is T/4 = 0.5 s, so 2.80 s holds 5 of them, plus 0.30 s.
Putting the angle in degrees into the formula with ω in rad/s. φ = 0.3π rad = 54°.
Mixing ω (rad/s) with the frequency f (Hz). Here ω = π, so f = 0.5 Hz.
The law is x = (35/π) sin πt, so a = 35/π cm and T = 2 s. In 2.80 s the point makes n = 5 quarter periods plus 0.30 s, and the distance is s = (35/π)(6 − cos 54°) ≈ 60 cm. For your sliders a = cm, n = , φ = degrees and s = cm.
Problem statement
At the moment t = 0 a particle starts moving along the x axis so that its velocity projection varies as vx = 35 cos πt cm/s, where t is in seconds. Find the distance that this particle covers during t = 2.80 s after the start.
We assume the particle starts at x = 0 at t = 0. This means the position is the integral of vx, and only the distance matters for the answer, so the start position does not change it.
We assume the motion goes on exactly as vx = v0 cos ωt for all time. This means a and ω stay fixed, with v0 = 35 cm/s and ω = π rad/s.
We assume "distance" means the total path length. This means the trip back counts as well as the trip out.
Symbols: v0 is the speed constant, ω the angular frequency, a = v0/ω the amplitude, T = 2π/ω the period, n the number of whole quarter periods in t, and φ the leftover phase.
The big idea in plain words
A velocity of the form v0 cos ωt belongs to a harmonic oscillation about x = 0. Integrating once gives the position law and shows the amplitude is v0/ω. Then the problem is the one we know: every quarter of a period covers exactly a, and only the leftover needs a formula. This time the particle starts at the middle, where the velocity is largest.
Interactive derivation
The solution in each frame
Lab frame
The axis is fixed to the lab, with x = 0 at the equilibrium position. The full solution is in the next section.
Point rest frame
This frame does not apply. The point keeps accelerating, so a frame fixed to it is not inertial, and it does not help to find the path length.
The solution step by step
First we find the motion law. The given velocity is vx = 35 cos πt, which has the form vx = aω cos ωt for x = a sin ωt. So we read off ω and a.
ω = π s⁻¹, a ω = 35 cm/s → a = 35/π cm ≈ 11.14 cm
T = 2π/ω = 2 s, T/4 = 0.5 s
Now we write the time in quarter periods. We need the whole number of quarters and what is left over.
t = 2.80 s = 5 × (T/4) + 0.30 s → n = 5, t0 = 0.30 s
φ = ω t0 = π × 0.30 = 0.3π = 54°
The particle starts at the middle, because x = a sin ωt is zero at t = 0. After an odd number of quarters it is at an extreme (n = 1 is the extreme x = a). So with n = 5 the leftover starts at an extreme, and it has moved from a down to a cos φ from there. Each full quarter covers exactly a.
s = n a + a(1 − cos φ) = a [n + 1 − cos φ]
s = (35/π)(5 + 1 − cos 54°) = 11.14 × (6 − 0.5878) = 11.14 × 5.412 = 60.3 cm ≈ 60 cm
When n is even the leftover starts at the middle and the formula changes to s = a[n + sin φ]. The page picks the right case by itself. For your sliders: a = cm, n = , so n a = cm, the leftover is cm, and s = cm.
Book check
Book answer: about 60 cm. The live value for the book setting is 60.3 cm, which matches to the book's rounding. The PDF has several typos in this solution: "x = a shoot" for x = a sin ωt, "(35/22) × 7" for 35/π with π taken as 22/7, and "interval 0 = 3 s", which should read 0.3 s. Its result 5 × (35/π) + (35/π)[1 − cos 0.3π] is the same as ours. A second formula (the area under the speed graph) gives the same distance, and a numeric integral of |vx| over the window agrees: cm against cm. A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
The law vx = v0 cos ωt gives wx = −v0 ω sin ωt = −ω² x, which is a spring-like force. So the motion is simple harmonic and the amplitude formula a = v0/ω holds. Force reasoning gives the type of motion, but the distance still needs the quarter-period count.
2. Energy method
At the middle the speed is the top speed v0, and at the ends it is zero. Energy gives (1/2) m v0² = (1/2) m ω² a², so a = v0/ω, the same result. It gives the amplitude quickly but not the distance for a given time.
3. Graph method
The distance is the area under the speed against time graph. Each quarter period has the same area a, so the distance graph climbs by a at each dotted line. The leftover is a part of one more hump.
4. Calculus method
This is the step-by-step route: s = ∫ |vx| dt with |vx| = 35 |cos πt|. Over each quarter the integral is a, and the leftover is a sin φ or a(1 − cos φ).
5. Check method
Units: a is a length, so s is a length. Limits: at t = T/4 the formula gives s = a, at t = T it gives 4a, as it must. For small t the distance is about v0 t, because the particle starts at its top speed. The distance never decreases, and it is never less than the displacement.
Fastest: the graph method. Most reliable: the calculus method, with the numeric check.
What each slider changes
Control
Changes the answer?
What it changes
Speed constant v0
Yes, proportional
The amplitude a = v0/ω, every distance, the track, the tape, the speed graph height
Angular frequency ω
Yes
a = v0/ω, T, the number of quarters n in t, the leftover phase, the distance and the motion speed
Time t
Yes
n, the leftover phase, the leftover distance, the total s and the tape
Screen speed
No
Only how fast the animation runs compared with real time
Target distance (given-answer mode)
Sets the answer
Solves t from s by bisection
Zoom, Auto-fit
No
Pixels per centimetre in the picture only
Parameters
Quantity
Symbol
Value used
Speed constant
v0
cm/s (book: 35)
Angular frequency
ω
rad/s (book: π)
Amplitude
a = v0/ω
cm
Period
T
s
Time
t
s (book: 2.80)
Whole quarter periods
n
Leftover phase
φ
degrees
Distance
s
cm
Exam traps
Using x(t) − x(0) as the distance. That is the displacement, which can even be zero after a full period.
Using the wrong leftover for n odd. After an odd number of quarters the point is at the middle, so the leftover is a sin φ, not a(1 − cos φ).
Counting n from half periods instead of quarter periods. A quarter period is T/4, not T/2.
Forgetting that the phase φ = ωt − nπ/2 is the leftover only, not the whole phase ωt.
Putting the angle in degrees into a formula with ω in rad/s. Keep radians.
Mixing ω (rad/s) with the frequency f (Hz). ω = 2π/T.