At 0.100 s intervals, the x
coordinates of a particle are
0, 1.01 cm, 2.05 cm, and 3.12
cm. Estimate the
instantaneous velocity and
acceleration at t = 0.
If you really want an expert to help you, be sure to follow directions and type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.
[f(.1)-f(0)]/.1 = .1
so, f'(0) is about .1
[f(.1)-f(0)]/.1 = .1
[f(.2)-f(.1)]/.1 = .5
so, f''(0) is about .4
looks like f(x) = 10x + x^2 + 4/3 x^3 ...
To estimate the instantaneous velocity and acceleration at t = 0, we can use the given x-coordinates of the particle at 0.100 s intervals.
To find the instantaneous velocity at t = 0, we need to calculate the average velocity between the first two data points (0 cm and 1.01 cm).
Average velocity = (change in position) / (change in time)
Change in position = 1.01 cm - 0 cm = 1.01 cm
Change in time = 0.100 s
Average velocity = 1.01 cm / 0.100 s = 10.1 cm/s
Therefore, the estimated instantaneous velocity at t = 0 is approximately 10.1 cm/s.
To find the instantaneous acceleration at t = 0, we need to calculate the average acceleration between the first three data points (0 cm, 1.01 cm, and 2.05 cm).
Average acceleration = (change in velocity) / (change in time)
Change in velocity = (1.01 cm/s - 10.1 cm/s) = -9.09 cm/s (note the negative sign)
Change in time = 0.100 s
Average acceleration = -9.09 cm/s / 0.100 s = -90.9 cm/s²
Therefore, the estimated instantaneous acceleration at t = 0 is approximately -90.9 cm/s² (negative sign indicates acceleration in the opposite direction).
Note: These calculations provide an estimate of the instantaneous velocity and acceleration at t = 0 based on the given data points. To obtain more precise values, additional data points or a more accurate mathematical model may be required.