Avg Acceleration problem + question on Speed vs Velocity

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The discussion revolves around calculating the average acceleration of a sprinter who accelerates for 4 seconds before maintaining maximum speed in a 100-meter race completed in 9.1 seconds. The correct average acceleration is determined to be 3.5 m/s², contrasting with an incorrect calculation of 2.75 m/s². Additionally, the conversation explores the difference between speed and velocity, clarifying that speed is the total distance traveled divided by time, while velocity considers displacement. The example of a 400-meter race illustrates how lane positions affect average velocity due to the curvature of the track. Overall, the distinction between speed and velocity is emphasized, particularly in races involving changes in direction.
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1. A world-class sprinter accelerates to his maximum speed in 4.0s. He maintains this speed for the remainder of the 100-m race, finishing with a time of 9.1 seconds.
Find A] What is the runners average acceleration during the first 4.0 seconds.




2. Vx=Vox+AxT
X=Xo+VoxT+1/2AxT
Vx2=Vox2+2Ax(X-Xo)
X-Xo=((Vox+Vx)/2)T




3. I used 100-m/9.1 seconds and got 10.99m/s as his average velocity. Divided it by 4 seconds to get an Acceleration of 2.75-m/s^2 but it is wrong. The correct answer is 3.5-m/s^2. Without knowing what distance he covered in the first 4 seconds I am stuck.

Also, I've been having a little trouble understanding speed vs velocity how can average speed be different than average velocity? I read in my text that speed is the magnitude of velocity and that it cannot be zero. Also, that "speed denotes distance traveled divided by time..." Which is exactly what Velocity equals; so how can they be different?
 
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resjsu said:
1. A world-class sprinter accelerates to his maximum speed in 4.0s. He maintains this speed for the remainder of the 100-m race, finishing with a time of 9.1 seconds.
Find A] What is the runners average acceleration during the first 4.0 seconds.




2. Vx=Vox+AxT
X=Xo+VoxT+1/2AxT
Vx2=Vox2+2Ax(X-Xo)
X-Xo=((Vox+Vx)/2)T




3. I used 100-m/9.1 seconds and got 10.99m/s as his average velocity. Divided it by 4 seconds to get an Acceleration of 2.75-m/s^2 but it is wrong. The correct answer is 3.5-m/s^2. Without knowing what distance he covered in the first 4 seconds I am stuck.

Also, I've been having a little trouble understanding speed vs velocity how can average speed be different than average velocity? I read in my text that speed is the magnitude of velocity and that it cannot be zero. Also, that "speed denotes distance traveled divided by time..." Which is exactly what Velocity equals; so how can they be different?


Firstly Velocity is displacement divided by time. Also in a straight race, like the 100m, where the runners do not change direction, there does appear to be very little difference between speed and velocity.

During steady acceleration, your average velocity is simply (initial vel + final vel) x 2

being a straight line event we could also say

During steady acceleration, your average speed is simply (initial speed + final speed) x 2

As you said your average speed is simply distance over time, and in this straight-line event the distance covered is the magnitude of your diplacement for the time.

[Note: compare that to the first 4 seconds of a 200m race where all the runners are running around the bend. The distance covered is an arc, the displacement is simply a chord to the circle]

Another [impossible] was to achieve the same average velocity would be to remain stationary for the 1st 2 seconds, then magically and mysteriously blast off at full speed for the next two seconds. so a 9.1 second 100m race is equivalent to a 7.1 second trip at full speed, [compared to the 4 second build up then the 5.1 second cruise]

There is some food for thought.

NB: in the 400m men's final in the London Olympics next year, the runner in the outside lane will have the largest average velocity.
Being the final - all runners will have a time within 5% of each other, but the staggered start means the runner in the outside lane starts the greatest distance from the finish line. Indeed, if they have a runner in Lane 1, that persons average velocity for the event will be 0

The winner will have the highest average speed!
 
Thank you for the reply but it is still confusing. Even in your reply you make it sound like speed and velocity are the same? Could you explain it a bit with the values in the problem above?

Also in a 400 meter doesn't everyone run... 400 meters?
 
Last edited:
resjsu said:
Thank you for the reply but it is still confusing. Even in your reply you make it sound like speed and velocity are the same? Could you explain it a bit with the values in the problem above?

Also in a 400 meter doesn't everyone run... 400 meters?

The 400m race best indicates the difference between speed and velocity.
We could best demonstrate it by having competitors run directly from their starting point, to their finishing point.

In the 100m sprint, that change would make no difference at all. Why - because the competitor never changes direction.

For the 400m race, it is very different.
Lane 1: start and finish point are one and the same place.
Lane 2: finish is about 8m behind the start point - around a bend.
Lane 3: finish is about 16m behind the start point - around a bend

For the marathon, some events the finish is several km from the start, but at the Olympics, where competitors often begin by running a few laps of the stadium before going out onto the streets only to return a couple of hours later and do one last lap of the stadium, the finish may also be the same place as the start [though often such races start on the opposite side of the Stadium to the finish].

In Australian Football, players have been tracked to find they run over 10km during the game - which lasts about 2 hours.
That would give an average speed of over 5 km/h
However the final position of the player may turn out to be the same as the starting position - standing in goal - so his average velocity for the game is zero.
Indeed the player with the highest average velocity will be the injured player taken to hospital [5 km away] during the first minute of the game!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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