i got a question in M1
N 03.....#7 (IV)
i solved almost 3/4 of it....but the final part doesnt wanna get right....so would some1 plz solve IV
i can use my knowledge of physics and additional mathas to solve this.
first find the (t) where velocity of B is maximum
for this differentiate velocity and equate to 0
d/dx (0.06t - 0.00012t
2) = 0.06 - 0.00024t
equate this to 0
0.06 - 0.00024t = 0
0.06 = 0.00024t
t = 0.06/0.00024 = 250s
now
at 250 seconds find the distance covered by B
for finding the distance integrate the velocity formula
<= integration symbol
<(0.06t - 0.00012t
2) = 0.03t
2 - 0.00004t
3 + c
when t = 0 ; s =0 therefore c = 0
so apply t=250 in the formula
s = 0.03t
2 - 0.00004t
3s = 1875 - 625 =
1250-
now find the distance traveled by A
u gotta use the diagram for this.
first find the equation of the line AB for this .
u wud get the equation
y = 0.012x + 3.6
substitute x=250 here
y = 6.6
name this point C
find the area of the trapezium ACXY
u wud get the area(distance) as 855
now find the area of the triangle AXO = 240
thus, the total area (distance) = 240 + 855 =
1095now 1250 - 1095 = 155m
hope that helped
find the diagram attached