# Thread: Calculate distance traveled velocity time graph: accel. uniformly to 30km/h in 2 min

1. ## Calculate distance traveled velocity time graph: accel. uniformly to 30km/h in 2 min

A driver starting a car from rest accelerates uniformly to a speed of 30km/h in 2 minutes.
She maintains this speed for another 5 minutes then applies the brakes and decelerates uniformly to rest in 1 minute.

Now they ask to draw a velocity time graph, no axes given, just a sketch, which I did.
Then they ask calculate the acceleration in km/h2 for the first two minutes which I did and it is 900km/h2
Then calculate the decelration in the last 2 minutes so it would be 0 -1800km/h2 = -1800km/h2

Now the part which is troubling me, they say calculate the total distance traveled in km. I know the area under the graph which is usually a trapezium is the total distance, however they just asked for a sketch, I don't really have any specific scales.
Their answer is 3(1/3)km but I'm not getting that.

I'm using this logic, for a period she traveled 5 minutes at 30km/h, so in 5 minutes the distance traveled would be 30km/h * 5/60 hours = 2.5 km.

Now I have to figure out the distance she traveled for the part where she accelerated and decelerated.. and yea I'm stuck.

2. Originally Posted by richiesmasher
A driver starting a car from rest accelerates uniformly to a speed of 30km/h in 2 minutes.
She maintains this speed for another 5 minutes then applies the brakes and decelerates uniformly to rest in 1 minute.

Now they ask to draw a velocity time graph, no axes given, just a sketch, which I did.
Then they ask calculate the acceleration in km/h2 for the first two minutes which I did and it is 900km/h2
Then calculate the decelration in the last 2 minutes so it would be 0 -1800km/h2 = -1800km/h2

Now the part which is troubling me, they say calculate the total distance traveled in km. I know the area under the graph which is usually a trapezium is the total distance, however they just asked for a sketch, I don't really have any specific scales.
Their answer is 3(1/3)km but I'm not getting that.

I'm using this logic, for a period she traveled 5 minutes at 30km/h, so in 5 minutes the distance traveled would be 30km/h * 5/60 hours = 2.5 km.

Now I have to figure out the distance she traveled for the part where she accelerated and decelerated.. and yea I'm stuck.
First you need to convert 30 km/hr → 0.5 km/min.

The small parallel side = 5

The large parallel side = 8

The height = 0.5

What is the area?

However, that answer is incorrect!

3. What is the difficulty with calculating the area of the remaining shapes?

4. Originally Posted by lev888
What is the difficulty with calculating the area of the remaining shapes?
My problem was the conversions I suppose, I was still using per hour, so I ended up having to use logic thinking I couldn't do it geometrically.

For instance, I kept thinking, I could do the area but my problem was not converting properly all along it seems.

5. Originally Posted by richiesmasher
My problem was the conversions I suppose, I was still using per hour, so I ended up having to use logic thinking I couldn't do it geometrically.

For instance, I kept thinking, I could do the area but my problem was not converting properly all along it seems.
Ok, I think I understand. In order to calculate the total area you don't need an exact sketch. You need to know what kind of shapes you are dealing with and the relevant dimensions. So, what are the shapes? What are the formulas for their areas? Do you have the needed dimensions?

6. Originally Posted by lev888
Ok, I think I understand. In order to calculate the total area you don't need an exact sketch. You need to know what kind of shapes you are dealing with and the relevant dimensions. So, what are the shapes? What are the formulas for their areas? Do you have the needed dimensions?
Yes in Mr khan's post he showed me what I did wrong, I knew the shape was a trapezium, and the area would be = (a+b) *h/2

But what I was doing wrong was putting the height as 30km per HOUR where it should have been minutes, and (a) and (b) are in minutes themselves.

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