# Math Help - Area Under A Graph

1. ## Area Under A Graph

Given that the total distance travelled by a train is $4 km$ and the deceleration is $4 ms^{-2}$ for the first $T$ second as shown in the speed-time graph below.Find:

a)the value of $T$

b)the value of $V$

2. The area under the curve represents that total distance traveled
So

$\frac{(100-V)T}{2} + VT + VT + 2VT + \frac{(100-V)\cdot 2T}{2} = \color{red}{4000}$ ........(and not 4)

And we are given deceleration so we get second equation

$100-4t= V$

Do you need more help?

The area under the curve represents that total distance travelled
So

$\frac{(100-V)T}{2} + VT + VT + 2VT + \frac{(100-V)\cdot 2T}{2} = 4$

And we are given decelaration so we get second equation

$100-4t= V$

Do you need more help?

sir,why area of trapezium sir do like 1/2(100-v)T?not 1/2(100+v)T?ur anwer is correct sir..but i want to understand..

4. First of all its not a trapezium .. (to know more about trapezium try to search the word "trapezium" in Wikipedia/google)

Ok here I will show you how I found that area

1st step:

Area of the two triangles (on the upper side) which will be formed when we form a line y= V

they are found using the formula 1/2 x (base) x (height)

Area of first triangle = 1/2 x (T) x (100-V)

Area of second triangle = 1/2 x (2T) x (100-V)

2nd Step :

Area of that rectangle which is below the line Y= V

Area of rectangle = V x 4T
-------------------------------
Add them you get the complete area

as in previous post

Feel free to ask if there is any other doubt.

5. thanks very much sir..and when i get the coordinates (0,100) and (20,v)..i solve the question b that is to find the value of v..i find it gradient like 100-v/0-20=4 and i get v =180?? but the answer is 20ms^-2..can u help me sir?

6. Use the second equation

100-4xT = V

This will give you the answer

V = 100-4x20

V =100-80 = 20

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Sorry for my mistake in first post I should have been more careful.
Its corrected in red.