Thread: Area Under A Graph

Given that the total distance travelled by a train is and the deceleration is for the first second as shown in the speed-time graph below.Find:

a)the value of

b)the value of

The area under the curve represents that total distance traveled
So

........(and not 4)

And we are given deceleration so we get second equation



Do you need more help?

Adarsh

Originally Posted by ADARSH

The area under the curve represents that total distance travelled
So



And we are given decelaration so we get second equation



Do you need more help?

Adarsh

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..

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

= Length x Breadth

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.

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?

Use the second equation

100-4xT = V


This will give you the answer

V = 100-4x20

V =100-80 = 20


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