A race car moves such that its position fits the relationship x = (5m/s) t + (0.
ID: 1759462 • Letter: A
Question
A race car moves such that its position fits the relationship x = (5m/s)t + (0.75m/s3)t3 where x is measured in meters and t inseconds. Determine the instantaneous velocity of the car att = 4.2 s, by computing theaverage velocity using a time interval of 0.40 s (i.e, in thiscase, the average velocity from 4 s to 4.4 s. Repeat for timeintervals of .20 s, and .10 s. In order to better see the limitingprocess keep exactly five significant figures in youranswer).t = 0.40 s t = 0.20 s t = 0.10 s
C. What is the average velocity at 4.2s? I have tried to complete this problem and I am apparently notgetting it so an answer and work shown will be lifesaver!!=]] x = (5m/s)t + (0.75m/s3)t3 where x is measured in meters and t inseconds. Determine the instantaneous velocity of the car att = 4.2 s, by computing theaverage velocity using a time interval of 0.40 s (i.e, in thiscase, the average velocity from 4 s to 4.4 s. Repeat for timeintervals of .20 s, and .10 s. In order to better see the limitingprocess keep exactly five significant figures in youranswer).
t = 0.40 s t = 0.20 s t = 0.10 s t = 0.40 s t = 0.20 s t = 0.10 s
C. What is the average velocity at 4.2s? I have tried to complete this problem and I am apparently notgetting it so an answer and work shown will be lifesaver!!=]] t = 0.40 s t = 0.20 s t = 0.10 s
Explanation / Answer
x = (5m/s)t + (0.75m/s3)t3 where x is measured in meters and t inseconds.Now the speed of the car is v = dx / dt
= (5m/s) + 3(0.75m/s3)t2
Now the instantaneous speed of thecar at t = 4.2s is
v = (5m/s) + 3(0.75 m/s3)(4.2s)2
= 44.7 m/s
Distance covered by the car at t = 4s is x = (5m/s)(4s) +(0.75m/s3)(4s)3
= 68m
Distance covered by the car at t = 4.4s is x = (5m/s)(4.4s) +(0.75 m/s3)(4.4s)3
= 85.888m
Now the average velocity of teh car in the time interval t = 4s tot = 4.4s is
v = (85.888m - 68m) /(4.4s - 4s)
= 44.72m/s