Question

A race car moves such that its position fits the relationship

*x* = (5.5 m/s)*t* + (0.60
m/s^{3})*t*^{3}

where *x* is measured in meters and *t* in
seconds.

(b) Determine the instantaneous velocity of the car at
*t* = 4.2 s, using time intervals of 0.40 s, 0.20 s, and
0.10 s. (In order to better see the limiting process *keep at
least three decimal places in your answer*.)

Δ*t* = 0.40 s . (???????) . m/s (Use the
interval from *t* = 4.00 s to 4.40 s.)

Δ*t* = 0.20 s . (???????) . m/s (Use the
interval from *t* = 4.10 s to 4.30 s.)

Δ*t* = 0.10 s (???????) . m/s (Use the interval from
*t* = 4.15 s to 4.25 s.)

(c) Compare the average velocity during the first 4.2 s with the results of part (b).

The average velocity of (???????) . m/s is (choose one (much less than - much greater than - about the same )?????)much less than about the same as much greater than the instantaneous velocity.

Answer #1

x = (5.5 m/s)t + (0.60 m/s3)t3

v = d(x)/dt

v = 5.5 + 1.8*t^2

x(4.4) = (5.5 m/s)*4.4 + (0.60 m/s3)*(4.4)^3 = 75.31

x(4.0) = (5.5 m/s)*4.0 + (0.60 m/s3)*(4.0)^3 = 60.4

x(4.3) = (5.5 m/s)*4.3 + (0.60 m/s3)*(4.3)^3 = 71.35

x(4.1) = (5.5 m/s)*4.1 + (0.60 m/s3)*(4.1)^3 = 63.90

x(4.25) = (5.5 m/s)*4.25 + (0.60 m/s3)*(4.25)^3 = 69.43

x(4.25) = (5.5 m/s)*4.15 + (0.60 m/s3)*(4.15)^3 = 63.90

(b)

x(4.2) = (x(4.4) - x(4.0))/0.4 = (75.31 - 60.4)/0.4 = 37.275

x(4.2) = (x(4.3) - x(4.1))/0.2 = (71.35 - 63.90)/0.2 = 37.25

x(4.2) = (x(4.25) - x(4.15))/0.1 = (69.43 -65.71)/0.1 = 37.2

(c)

v = 5.5 + 1.8*t^2

v(4.2) = 5.5 + 1.8*4.2^2 = 37.252 m/s

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