Answered

IDNLearn.com makes it easy to find accurate answers to your specific questions. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.

During the time interval from 0.0 to 10.0 s, the position vector of a car on a road is given by x(t) = a + bt + ct2, with a = 17.2 m, b = -10.1 m/s, and c = 1.10 m/s2. What is the car’s velocity as a function of time? What is the car’s average velocity during this interval?

im stuck with this :(

Sagot :

Onyebuchinnajiidn

The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

Average velocity of the car

The average velocity of the car is calculated as follows;

x(t) = a + bt + ct2

v = dx/dt

v(t) = b + 2ct

v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s

v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s

Average velocity

V = ¹/₂[v(0) + v(10)]

V = ¹/₂ (-10.1  + 11.9 )

V = 0.9 m/s

Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

Learn more about velocity here: https://brainly.com/question/4931057

#SPJ1

Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.