a 1300-kilogram space vehicle travels at 4.8 meters per second along the level surface of mars. if the magnitude of the gravitat
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The average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds = 63.84 ft/s
(ii) 0.001 seconds = 63.984 ft/s
Given that a ball is thrown with an initial velocity = 80 ft/s
Let 'y' be the height in feet after 't' seconds.
Given, gives the height in 't' seconds.
Average velocity = Rate of change of distance
= Change in distance/Change in time.
The initial time can be taken as 0 s.
When t =1 s, y = 80 - 16 = 64 ft
(1) t = 0.01 s
y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft
Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s
(2) t = 0.001 s
y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft
Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s
The question is incomplete. Find out the complete question below:
If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by .Find the average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds
(ii) 0.001 seconds
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Answer:
The acceleration is a = 2.75 [m/s^2]
Explanation:
In order to solve this problem we must use kinematics equations.
where:
Vf = final velocity = 13 [m/s]
Vi = initial velocity = 2 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Now replacing:
13 = 2 + (4*a)
(13 - 2) = 4*a
a = 2.75 [m/s^2]