Question

A race car moves such that its position fits the relationship

x = (5.5 m/s)t + (0.60 m/s³)t³

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.)

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

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

Δt = 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