an object starts from rest and travels 100 m in 10 s. what will be the velocity of the object after 10 s
1 answer:
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Refer to the diagram shown below.
When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m
Answer: The maximum height attained is 24.7 m (nearest tenth)
Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s
To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.
Answer: The ball stays in the air for 4.5 s (nearest tenth)
When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m
Answer: The maximum height attained is 24.7 m (nearest tenth)
Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s
To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.
Answer: The ball stays in the air for 4.5 s (nearest tenth)
The complete text of the problem is:
<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>
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<u>Solution:</u>
1) Acceleration: on the hardwood floor,
on the carpeted floor
First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation
where
v is the final speed
u = 0 is the initial speed (the child starts from rest)
a = g = 9.8 m/s^2 is the acceleration of gravity
d = 0.43 m is the distance covered by the child as he falls from the bed
Solving for v,
Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is
So we can find the acceleration by using again the equation
Solving for a,
For the carpeted floor instead,
therefore the acceleration is
2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor
We can find the duration of the collision in both cases by using the equation of the acceleration
where
v' = 0
v = 2.9 m/s
For the hardwood floor,
So the duration of the collision is
For the carpeted floor,
So the duration of the collision is
We can now comment the results using the initial statement of the problem:
"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"
Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).