# Acceleration and Distance

1. Feb 17, 2008

### razored

[SOLVED] Acceleration and Distance

1. The problem statement, all variables and given/known data
"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?"

2. Relevant equations
d = .5 a t^2 + Vi* t
a = V/t

2. Feb 17, 2008

### Kurdt

Staff Emeritus
For the first part it would be best to used the average speed to work out the distance.

$$s =\left(\frac{u+v}{2}\right) t$$

3. Feb 17, 2008

### _Mayday_

Hey Razored.

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.

4. Feb 17, 2008

### HallsofIvy

Staff Emeritus
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. Feb 17, 2008

### razored

I got 34m total, is that correct?

6. Feb 17, 2008

### Staff: Mentor

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. Feb 17, 2008

### Staff: Mentor

Yep.

8. Feb 17, 2008

### razored

Thanks.

9. Feb 17, 2008

### Staff: Mentor

10. Feb 17, 2008

### Kurdt

Staff Emeritus
Its amazing how much service one can drum up in the chat room.

11. Feb 17, 2008

### _Mayday_

^ Haha I know! The question is also nice to do, or atleast I enjoyed it nice a simple. =]