Consider a car whose position, s, is given by the table
t(s) s(ft)
0 0
.2 0.35
.4 1.8
.6 3.8
.8 6.5
1 9.6
Estimate the velocity at t=0.2
t(s) s(ft) Δs Δ²s
0 0 0.35 1.1
.2 0.35 1.45 0.55
.4 1.8 2 0.7
.6 3.8 2.7 0.4
.8 6.5 3.1
1 9.6
The table above was completed using forward difference.For example, 0.35 = 0.35-0.0, 1.45=1.8-0.35, etc. Similarly for the fourth column.
The acceleration is not uniform, as evidenced by the fourth column, which is an indication of the acceleration (decreasing).
An estimate of the derivative can be made by the central difference formula, namely:
f'(t)=(f(t+h)-f(t-h))/2h + ε(O h²)
When t=0.2, we get
f'(0.2)=(1.8-0.0)/(2*0.2)=4.5 f/s²
Using a different formula (forward difference),
f'(t)=(-3f(t)+4f(t+h)-f(t+2h))/(2h) + ε(O h²)
=(-3*0.35+4*1.8-3.8)/(2*0.2)
=5.88 f/s²
Most likely the value would lie between the two values 4.5 and 5.88 f/s².
Well, let's calculate the average velocity between t = 0 and t = 0.2. To do this, we'll use the formula:
velocity = (change in position) / (change in time)
So, the change in position between t = 0 and t = 0.2 is 0.35 ft - 0 ft = 0.35 ft. And the change in time is 0.2 - 0 = 0.2 s.
Now, divide the change in position by the change in time:
velocity = 0.35 ft / 0.2 s
And now, let's find our calculators and do the math! Just hold on one moment...
To estimate the velocity at t=0.2, we need to calculate the average velocity between t=0 and t=0.4.
Velocity is defined as the rate of change of position with respect to time. In this case, we can estimate the velocity by finding the change in position divided by the change in time within a small interval.
Using the table given, we can see that at t=0.2, the position is 0.35 ft and at t=0.4, the position is 1.8 ft.
To calculate the average velocity, we divide the change in position (1.8 - 0.35 = 1.45 ft) by the change in time (0.4 - 0 = 0.4 sec):
Average velocity = (1.8 - 0.35) ft / (0.4 - 0) sec
= 1.45 ft / 0.4 sec
= 3.625 ft/sec
Therefore, the estimated velocity at t=0.2 is approximately 3.625 ft/sec.