Given :
A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .
To Find :
A. The vector position of the particle at any time t .
B. The acceleration of the particle at any time t .
Solution :
A )
Position of vector v is given by :
B )
Acceleration a is given by :
Hence , this is the required solution .
Answer:
x = 24
Step-by-step explanation:
18 + 0.5x = 12 +0.75 x
18 - 12 = 0.75 x - 0.5x
6 = 0.25 x
24 = x
Answer:
s=P-b/2
Step-by-step explanation:
P-b=2s
P-b/2=s
Answer:
x = 38, y = 4
Step-by-step explanation:
Since AB = BC then the triangle is isosceles and the base angles are congruent, that is
∠ DAB = ∠ DCB = 52°
Subtract the sum of the base angle from 180° for ∠ ABC
∠ ABC = 180° - (52 + 52)° = 180° - 104° = 76°
Note that ∠ ABD = ∠ CBD, thus
x = 76 ÷ 2 = 38
BD bisects the side AC, thus DC = AD = 4
Thus y = 4
