A race car moves such that its position fits the relationship x = (4.0 m/s)t + (

Question

A race car moves such that its position fits the relationship x = (4.0 m/s)t + (0.80 m/s^3)t^3 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? (b) Determine the instantaneous velocity of the car at t = 4.1 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.) Delta t = 0.40 s m/s (Use the interval from t = 3.90 s to 4.30 s.) Delta t = 0.20 s m/s (Use the interval from t = 4.00 s to 4.20 s.) Delta t = 0.10 s m/s (Use the interval from t = 4.05 s to 4.15 s.) (c) Compare the average velocity during the first 4.1 s with the results of part (b). The average velocity of m/s is the instantaneous velocity.

Explanation / Answer

(a)

given equation is

x = 4*t + (0.8)*t3

at t =0 , x = 0

t = 1 , x = 4.8

t = 2 , x = 14.4

t = 3 ,x = 33.6

according to the above values graph third is right answer.

(b)

we know that

v = dx/dt

v = 4+2.4*t2

at t = 4.1 s

v = 4 + 2.4*4.1*4.1

v = 44.34 m/s

(c)

Distance covered by the car at t = 4s is

x = (4)*(4) +(0.8)*(4)3

x = 67.2 m

Distance covered by the car at t = 4.4s is

x = (4)*(4.4s) +(0.8 )*(4.4)3
x =    85.74m
Now the average velocity of the car in the time interval t = 4s to t = 4.4s is
v = (85.74 - 67.2) /(4.4s - 4s)
v = 46.36 m/s   

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