The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion
s = 2 sin(πt) + 2 cos(πt),
where t is measured in seconds. (Round your answers to two decimal places.)
(a) Find the average velocity during each time period.
(i) [1, 2]
cm/s
(ii) [1, 1.1]
cm/s
(iii) [1, 1.01]
cm/s
(iv) [1, 1.001]
cm/s
(b) Estimate the instantaneous velocity of the particle when t = 1.
cm/s
https://www.jiskha.com/questions/1813507/the-displacement-in-centimeters-of-a-particle-moving-back-and-forth-along-a-straight
careful, displacement not distance (vector)
lol, I did part iii too
By the way they luckily asked for displacement and velocity, not average distance and speed. If they asked for the scalars and not the vectors you would have to be careful about going back and forth. If you go all the way around a circular track your displacement and average velocity is ZERO. However your distance is pi D and speed is pi D/time
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Damon
today at 5:41pm
This problem is trying to get you to converge to the speed at t = 1
which we know is -2 pi
so try the part iii for example
s(1.01 ) = 2 sin(1.01 pi ) + 2 cos(1.01 pi)
= -0.0628 -1.999 = -2.062
s(1) = 2(0)+2(-1) = -2
so
s(1.01)-s(1) = -.062
divide by time of 0.01 seconds
v average = -6.1813
the exact from the second part was -2 pi = - -6.2318 .......
if you did 1 to 1.0001 seconds it would be even closer
To find the average velocity during different time periods, we need to calculate the displacement and divide it by the time interval.
(a) Average Velocity:
(i) Time period [1, 2]:
- The displacement during this time period can be found by subtracting the initial position (t = 1) from the final position (t = 2).
s_final = 2sin(π(2)) + 2cos(π(2)) = -2 cm
s_initial = 2sin(π(1)) + 2cos(π(1)) = 2 cm
Displacement = s_final - s_initial = -4 cm
- The time interval is 2 - 1 = 1 second.
- Average velocity = Displacement / Time interval = -4 cm / 1 s = -4 cm/s
(ii) Time period [1, 1.1]:
- The displacement during this time period can be found similarly.
s_final = 2sin(π(1.1)) + 2cos(π(1.1)) ≈ 2.99 cm
s_initial = 2sin(π(1)) + 2cos(π(1)) = 2 cm
Displacement = s_final - s_initial ≈ 0.99 cm
- The time interval is 1.1 - 1 = 0.1 second.
- Average velocity = Displacement / Time interval ≈ 0.99 cm / 0.1 s ≈ 9.90 cm/s
(iii) Time period [1, 1.01]:
- Repeat the same procedure.
s_final = 2sin(π(1.01)) + 2cos(π(1.01)) ≈ 2.039 cm
s_initial = 2sin(π(1)) + 2cos(π(1)) = 2 cm
Displacement = s_final - s_initial ≈ 0.039 cm
- The time interval is 1.01 - 1 = 0.01 second.
- Average velocity = Displacement / Time interval ≈ 0.039 cm / 0.01 s ≈ 3.90 cm/s
(iv) Time period [1, 1.001]:
- Repeat the same procedure.
s_final = 2sin(π(1.001)) + 2cos(π(1.001)) ≈ 2.0039 cm
s_initial = 2sin(π(1)) + 2cos(π(1)) = 2 cm
Displacement = s_final - s_initial ≈ 0.0039 cm
- The time interval is 1.001 - 1 = 0.001 second.
- Average velocity = Displacement / Time interval ≈ 0.0039 cm / 0.001 s ≈ 3.90 cm/s
(b) Instantaneous Velocity:
To estimate the instantaneous velocity when t = 1, we can find the derivative of the displacement equation with respect to time and evaluate it at t = 1.
The derivative of s with respect to t is:
s'(t) = 2πcos(πt) - 2πsin(πt)
Evaluate s'(t) at t = 1:
s'(1) = 2πcos(π(1)) - 2πsin(π(1)) ≈ -6.28 cm/s
Therefore, the estimated instantaneous velocity of the particle when t = 1 is approximately -6.28 cm/s.
I will do iii) you do the rest in the same way
s = 2 sin(πt) + 2 cos(πt)
[1, 1.01] cm/s
when t = 1
s = 2sinπ + 2cosπ = 0 - 2 = -2
when t = 1.01
s = 2sin(1.01π) + 2cos(1.01π) = -2.0618346...
avg velocity = change in distance/change in time
= (-2.0618346.. - (-2))/(1.01-1) = -6.183 or -6.18 correct to 2 decimals