Question

A race car moves such that its position fits the relationship x (6.0 m/s)t +(0.85 m/s3)t3 where x is measured in meters and t in seconds. (a) A plot of the car's position versus time is which of the following? x (m) 240 x (m) 240 220 200 18 16 12 14 60 r (s) t (s) 1.00 2.00 3.00 4.00 5.00 6.00 1.00 2.00 3.00 4.00 5.00 6.00 x (m) 30 25 x (m) 15 r (s) t (s) 1.00 2.00 3.00 4.00 5.00 6.00 1.00 2.00 3.00 4.00 5.00 6.00 (b) Determine the in s velocity of the car at t 3.3 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to better see the limiting process keep atleast three decimal places in your answer.) m/s (Use the interval from t 3.10 s to 3.50 s.) m/s (Use the interval from t 3.20 s to 3.40 s.) m/s (Use the interval from t3.25 s to 3.35 s.) ΔΙ-0.40 s At-0.20 s (c) Compare the average velocity during the first 3.3 s with the results of part (b) The average velocity of m/s is-Select- the instantaneous velocity

Answer 1

given

x = 6.0 m/s t + 0.85 m/s³ t³

a )

is correct representation.

b )

x = 6.0 m/s t + 0.85 m/s³ t³

then v = dx/dt

= 6 + 3 x 0.85 x t²

v = 6 + 2.55 t²

at t = 3.3 sec

v = 6 + 2.55 t²

= 6 + 2.55 x (3.3)²

= 6 + 27.7695

v = 33.7695 m/sec

c )

at t = 0 sec

displacement of the body is x = 0 m

at t = 3.3 sec

the displacement of the body is x = 6 x 3.3 + 0.85 x 3.3³

= 19.8 + 30.54645

x = 50.34645 m

then the average velocity is V_avg = 50.34645 / 3.3

V_avg = 15.2565 m/sec