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.
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
1 answer