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.
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C is the correct answer.
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<span>a) 1960 m
b) 960 m
Assumptions.
1. Ignore air resistance.
2. Gravity is 9.80 m/s^2
For the situation where the balloon was stationary, the equation for the distance the bottle fell is
d = 1/2 AT^2
d = 1/2 9.80 m/s^2 (20s)^2
d = 4.9 m/s^2 * 400 s^2
d = 4.9 * 400 m
d = 1960 m
For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes
d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T
d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s
d = 4.9 m/s^2 * 400 s^2 - 1000 m
d = 4.9 * 400 m - 1000 m
d = 1960 m - 1000 m
d = 960 m</span>
Answer:
0.9 N
Explanation:
The force exerted on an object is related to its change in momentum by:
where
F is the force exerted
is the change in momentum
is the time interval
The change in momentum can be rewritten as
where
m is the mass
u is the initial velocity
v is the final velocity
So the formula can be rewritten as
In this problem we have:
is the mass rate
is the initial velocity
is the final velocity
Therefore, the force exerted by the hail on the roof is:
