How far did the bike travel with acceleration and constant speed?

In summary, the problem involves a bike accelerating from 0.0m/s to 5.0m/s in 4.5secs and then continuing at a constant speed for another 4.5s. Using the average speed equation, the distance traveled during the first period is 11.25m. During the second period, the distance traveled is 22.5m. Thus, the total distance traveled by the bike is 33.75m.
  • #1
razored
173
0
[SOLVED] Acceleration and Distance

Homework Statement


"A bike first accelerates from 0.0m/s to 5.0m/s in 4.5secs, then continues at this constant speed for another
4.5s. What is the total distance traveled by the bike?"


Homework Equations


d = .5 a t^2 + Vi* t
a = V/t
 
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  • #2
For the first part it would be best to used the average speed to work out the distance.

[tex] s =\left(\frac{u+v}{2}\right) t [/tex]
 
  • #3
Hey Razored. :smile:

Ok, I am sure there is more than one way to approach this question. I would draw out a speed time graph, with time along the x-axis and time up the y-axis, then calculate the area under the graph. Do you understand that? If not just let me know and I will try and explain. These forums usually ask that you make or show some workings first. :smile:
 
  • #4
Okay, you have the equations, and you have the numbers. Now do the arithmetic! What is the acceleration if the bike accelerated from 0 to 5 m/s in 4.5 s? Just put the numbers into the second equation. Use the first equation to find the distance the bike traveled in that time.

The second part is easy. When the bike got to 5 m/s it stopped accelerating- it continued at 5 m/s for 4.5 s. How far did it go in the second 4.5 seconds?

Now find the total distance it travelled.
 
  • #5
I got 34m total, is that correct?
 
  • #6
razored said:
I got 34m total, is that correct?
Correct with round off.

As others mentioned there are two parts. One period of 4.5 s at constant acceleration and one period of 4.5 s at constant velocity.

During the first period, the average v is 2.5 m/s, so d = 2.5 m/s * 4.5 s, and in the second V = 5 m/s, so d = 5 m/s * 4.5 s.
 
  • #7
Yep.
 
  • #8
Thanks.
 
  • #10
Its amazing how much service one can drum up in the chat room.
 
  • #11
^ Haha I know! The question is also nice to do, or atleast I enjoyed it nice a simple. =]
 

1. What is acceleration?

Acceleration is the rate of change of velocity, or how quickly an object's speed is increasing or decreasing over time. It is measured in units of distance divided by time squared, such as meters per second squared (m/s²).

2. How is acceleration calculated?

Acceleration can be calculated using the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. This equation is derived from the definition of acceleration as the change in velocity divided by the change in time.

3. What is the relationship between acceleration and distance?

The relationship between acceleration and distance is indirectly proportional. This means that as acceleration increases, distance traveled also increases, and vice versa. This relationship is described by the equation d = vi*t + 1/2*a*t², where d is distance, vi is initial velocity, t is time, and a is acceleration.

4. How does mass affect acceleration?

Mass does not directly affect acceleration, but it does affect the amount of force needed to accelerate an object. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that as mass increases, more force is needed to achieve the same acceleration.

5. What is the difference between average and instantaneous acceleration?

Average acceleration is the overall change in velocity over a specific time interval, while instantaneous acceleration is the acceleration at a specific moment in time. In other words, average acceleration takes into account the entire motion of an object, while instantaneous acceleration only considers a specific point in time.

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