Question #46688

The position ) f (x of a particle moving in a line at various times k
x is given in the
following table. Estimate the velocity and acceleration of the particle at x = 2.1 .

x 1.0 1.2 1.4 1.6 1.8 2.0 2.2
f (x) 2.72 3.32 4.06 4.96 6.05 7.39 9.02
1

Expert's answer

2014-09-30T08:40:42-0400

Answer on Question #46688 – Math – Algorithms | Quantitative Methods

The position f(x)f(x) of a particle moving in a line at various times kk

xx is given in the following table. Estimate the velocity and acceleration of the particle at x=2.1x = 2.1.

x 1.0 1.2 1.4 1.6 1.8 2.0 2.2

f (x) 2.72 3.32 4.06 4.96 6.05 7.39 9.02

Solution.

Velocity at x=2.1x = 2.1: v(2.1)=f(2.2)f(2.0)2.22.0=9.027.390.2=8.15v(2.1) = \frac{f(2.2) - f(2.0)}{2.2 - 2.0} = \frac{9.02 - 7.39}{0.2} = 8.15.

Velocity at x=2.2x = 2.2: v(2.1)=f(2.2)f(2.0)2.22.0=9.027.390.2=8.15v(2.1) = \frac{f(2.2) - f(2.0)}{2.2 - 2.0} = \frac{9.02 - 7.39}{0.2} = 8.15.

Velocity at x=2.0x = 2.0: v(2.0)=f(2.0)f(1.8)2.01.8=7.396.050.2=6.70v(2.0) = \frac{f(2.0) - f(1.8)}{2.0 - 1.8} = \frac{7.39 - 6.05}{0.2} = 6.70.

Acceleration at x=2.1x = 2.1: a(2.1)=v(2.2)v(2.0)2.22.0=8.156.700.2=9.00a(2.1) = \frac{v(2.2) - v(2.0)}{2.2 - 2.0} = \frac{8.15 - 6.70}{0.2} = 9.00.

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