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2 years ago

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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 :(

1 answer:

Juli2301 [7.4K]2 years ago

3 0

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.

<h3>Average velocity of the car</h3>

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

<h3>Average velocity</h3>

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: brainly.com/question/4931057

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