# The velocity v(t) of an object is sketched below. Estimate the total distance travelled by an object between t= 0 and t= 4The velocity v(t) of an object is sketched below. Estimate the...

The velocity v(t) of an object is sketched below. Estimate the total distance travelled by an object between t= 0 and t= 4

The velocity v(t) of an object is sketched below. Estimate the total distance travelled by an object between t= 0 and t= 4

link to graph: http://s19.postimage.org/mkf7knvar/calc_hw_21_graph_for_6.png

[url=http://postimage.org/image/vs7g1d2cv/][img]http://s19.postimage.org/vs7g1d2cv/calc_hw_21_graph_for_6.jpg[/img][/url]

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The area covered below the graph gives you the total travel distance.

Usually graphs like this are integrated to get the area. But we can integrate it if we know the equation of the graph.

In this case there is no equation is given. So we cannot integrate for the area.

Since the question says to estimate the area we can use approximations for finding area.

If you see the graph we can see from t=0 to t=3 the velocity vs.time graph has curve shape. But from t=3 to t=4 it looks like a straight line.

So i use two methods to obtain the area of each curve and straight parts as stated earlier.

From t=0 to t=3

The area below the curve = Numbers of squares below the curve.

Here I consider a** small square** below curve where area is 1*1 = 1

- If more than 1/2 of the square is below the curve, consider it as 1 square
- If less than 1/2 of the square is below the curve it will be omitted.
- If a square is fully under the curve it is also considered as 1 square.

Squares more that 1/2 under curve = 2

squares fully under the curve = 1

Total area under curve= 2+1 =3

From t=3 to t=4

Here we have a trapiziod.

Area under straight line = (1.7+5.7)/2*1

= 3.7

So total area under the grapgh = 3+3.7 = 6.7

As stated earlier the area of grapgh = Distance travelled by object.

**So distance travelled by object = 6.7**