A particle moves in one dimension, and its position as a function of time is given by
x = (1.7 m/s)t + (−3.0 m/s2)t2.
(a) What is the particle's average velocity from t = 0.45 s to t = 0.55 s? (Indicate the direction with the sign of your answer.)
(b) What is the particle's average velocity from t = 0.49 s to t = 0.51 s? (Indicate the direction with the sign of your answer.)
To find the average velocity of the particle between two given time intervals, we need to calculate the displacement of the particle during that time and divide it by the time taken.
(a) To find the average velocity from t = 0.45 s to t = 0.55 s, we need to find the displacement of the particle during this time interval. The equation given for the position of the particle as a function of time is:
x = (1.7 m/s)t + (-3.0 m/s^2)t^2
We can find the displacement by subtracting the initial position from the final position. Let's calculate:
At t = 0.45 s:
x1 = (1.7 m/s)(0.45 s) + (-3.0 m/s^2)(0.45 s)^2
= 0.765 m - 0.6075 m
= 0.1575 m
At t = 0.55 s:
x2 = (1.7 m/s)(0.55 s) + (-3.0 m/s^2)(0.55 s)^2
= 0.935 m - 0.8925 m
= 0.0425 m
Displacement = x2 - x1
= 0.0425 m - 0.1575 m
= -0.115 m
The negative sign indicates that the particle is moving in the opposite direction. Therefore, the average velocity from t = 0.45 s to t = 0.55 s is -0.115 m divided by the time interval, which is 0.55 s - 0.45 s = 0.1 s.
Average velocity = Displacement / Time interval
= -0.115 m / 0.1 s
= -1.15 m/s
Therefore, the particle's average velocity from t = 0.45 s to t = 0.55 s is -1.15 m/s (in the opposite direction of the positive x-axis).
(b) To find the average velocity from t = 0.49 s to t = 0.51 s, we follow the same process.
At t = 0.49 s:
x1 = (1.7 m/s)(0.49 s) + (-3.0 m/s^2)(0.49 s)^2
= 0.833 m - 0.7153 m
= 0.1177 m
At t = 0.51 s:
x2 = (1.7 m/s)(0.51 s) + (-3.0 m/s^2)(0.51 s)^2
= 0.867 m - 0.7653 m
= 0.1017 m
Displacement = x2 - x1
= 0.1017 m - 0.1177 m
= -0.016 m
The negative sign indicates that the particle is moving in the opposite direction. The time interval is 0.51 s - 0.49 s = 0.02 s.
Average velocity = Displacement / Time interval
= -0.016 m / 0.02 s
= -0.8 m/s
Therefore, the particle's average velocity from t = 0.49 s to t = 0.51 s is -0.8 m/s (in the opposite direction of the positive x-axis).
To find the particle's average velocity, we need to calculate the displacement and divide it by the time interval.
(a) Let's find the particle's displacement from t = 0.45 s to t = 0.55 s.
The position function is given by x = (1.7 m/s)t + (-3.0 m/s^2)t^2.
At t = 0.45 s:
x1 = (1.7 m/s)(0.45 s) + (-3.0 m/s^2)(0.45 s)^2
x1 = 0.765 m - 0.6075 m
x1 = 0.1575 m
At t = 0.55 s:
x2 = (1.7 m/s)(0.55 s) + (-3.0 m/s^2)(0.55 s)^2
x2 = 0.935 m - 0.98475 m
x2 = -0.04975 m
The displacement from t = 0.45 s to t = 0.55 s is:
Δx = x2 - x1
Δx = -0.04975 m - 0.1575 m
Δx = -0.20725 m
The time interval is:
Δt = t2 - t1
Δt = 0.55 s - 0.45 s
Δt = 0.1 s
The average velocity is given by:
Average velocity = Δx / Δt
Average velocity = -0.20725 m / 0.1 s
Average velocity ≈ -2.0725 m/s
Therefore, the particle's average velocity from t = 0.45 s to t = 0.55 s is approximately -2.0725 m/s, in the negative direction.
(b) Let's find the particle's displacement from t = 0.49 s to t = 0.51 s.
Using the same position function, we can calculate the position at these times.
At t = 0.49 s:
x1 = (1.7 m/s)(0.49 s) + (-3.0 m/s^2)(0.49 s)^2
x1 = 0.833 m - 0.72471 m
x1 = 0.10829 m
At t = 0.51 s:
x2 = (1.7 m/s)(0.51 s) + (-3.0 m/s^2)(0.51 s)^2
x2 = 0.867 m - 0.77577 m
x2 = 0.09123 m
The displacement from t = 0.49 s to t = 0.51 s is:
Δx = x2 - x1
Δx = 0.09123 m - 0.10829 m
Δx ≈ -0.01706 m
The time interval is:
Δt = t2 - t1
Δt = 0.51 s - 0.49 s
Δt = 0.02 s
The average velocity is given by:
Average velocity = Δx / Δt
Average velocity = -0.01706 m / 0.02 s
Average velocity ≈ -0.853 m/s
Therefore, the particle's average velocity from t = 0.49 s to t = 0.51 s is approximately -0.853 m/s, in the negative direction.