Scientists use specific processes to evaluate information in order to determine its accuracy and reliability. A scientist is con
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a) 4.8 m/s² his average acceleration during the first 2.5 s
b) 15 m distance he cover during the first 2.5 seconds.
c) 31 m/s his speed as he finished the race.
<h3>(a) How to calculate the Average Acceleration ?</h3>Average acceleration is calculated by using Newtons first law of motion
v = u + at
where
u is initial velocity
v is final velocity
a is constant acceleration
t is time
Given u = 0m/s , v = 12 m/s and t = 2.5 sec
Therefore average acceleration is given by
a = 4.8 m/sec²
b) Here we will use newtons second law of motion
On substituting value we get
s = 15m
c) Here we will use newtons third law of motion
v² = u² + 2as
Here u = 0 , a = 4.8 m/s² and s = 100 m
Therefore
v² = 960
v = 31 m/s
Disclaimer: the question was given incomplete in the portal. Here is the complete question.
Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.
a. What was his average acceleration during the first 2.5 s?
b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?
c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?
Learn more about Newtons law of motion here:
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Hope this provides some answers
Answer:
Explanation:
When the ball is at the bottom position of the vertical circle, the forces acting on the ball are:
- The tension in the string, , upward
- The weight of the ball, , downward
The resultant of these forces must be equal to the centripetal force, which points upward as well (towards the center of the circle), so:
(1)
where is the speed of the ball at the bottom of the circle, r the radius of the circle, and m the mass of the ball.
When the ball is at the top position of the vertical circle, the weight still acts downward, however the tension in the string also acts downward, so the equation of the forces becomes:
(2)
where
is the tension in the string in the top position
is the speed of the ball in the top position
By subtracting eq.(2) from eq.(1), we find:
So, this is the difference in tension between the two positions.