1. A motorboat going Downstream overcame a raft at a point A ; T= 60 min later it turned back and after sometime passed the raft at a distance l = 6.0 km from the point A. Find the velocity,assuming the duty of the engine to be constant.
Answer : Upstream - Downstream motion has to be calculated as per Time.
Unlike Distance based calculations which are two-dimensional ,Time based calculations are made as three-dimensional which includes the direction vector as well
At first, Dimensional Analysis is done to find out the equation strategy. This has been called dimensional analysis of vector equations . This is done as follows:
Since distance based calculation is involved ,
distance / time = velocity ,which is a vector
Hence [VT] = a constant (distance)
Hence the length between ' B' and 'P' in the figure ,which is 'BC' has to be found out .
3-Dimensional approach to this 2-d problem is drawn below
Time = Distance / Velocity ;
Therefore, (l-BC)/Velocity raft = AP / Velocity P = 60 min
That means, (6 - BC )/V raft = 60 minutes = 1hour
V raft = (6 - BC) / 1
As per modern trigonometry,
( Hypotnuese - square ) =( Base - square) + ( Altitude -square)
As per modern Trigonometry, Hypotnuese is 2 times bigger than altitude and base for unity triangle ( which is 1 mm in base )
Now, Indian Mathematics suggests ,PASCAL'S TRIANGLE for solving Trigonometry problems that are unknown to calculation.
Here ,1 + 4 = 5 , but 6 is given ,Hence, we are showing it as given above in Pascals image .
Now if you calculate diagonally, these numbers can introduce you to Trigonometry ,which is as shown below
1 4 5 9 5 , which is known as Pascal's rule in trigonometry. ( Inclined indicates triangular formation using number digits.)
So, 1 (square ) + 4 ( square ) = square root ( 17) which indicates that square has to be used diagonally in the Pascal's format shown above.
So , 1 2 √3 in triangle calculation becomes 1 2 3 as shown in Pascal's Triangle above ( diagonally)
So, Above BC becomes 3 ( And that is the Solution )
