If an object starts at rest and moves 60 meters north along a straight line in 2 seconds, what is the average velocity?
2 answers:
Answer:
The average velocity, expressed as a vector, is v=30 m/s with direction north.
Explanation:
In this case, we will obtain a vector velocity, and <u>the correct answer will be expressed as a vector</u>. But as <em>the movement happens only along a straight line</em> (with direction to the north from the point of rest), we can do the math calculation as a <u>one-dimensional problem</u>. Then <em>the final velocity will be a vector with the calculated magnitude, pointing to the north</em>, as we already know.
The question is simple, as we were given the distance in meters that the object moved in a straight line, and the time in seconds that took the object to move that distance, we have the correct units and we can write the average velocity as
where d is the distance, and t is the time. Therefore
<u>which is the magnitude of the velocity we wanted to know</u>, then <em>we express it as vector saying its direction (north, for our system of reference)</em>.
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Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.
The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.
where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]