The position vector for particle a is cos(t)i, and the position vector for particle b is sin(t)j. what is the difference in acceleration (i. e. the relative acceleration) between particle a and b at any time t? the acceleration vector of a particle moving in space is the second derivative of the position vector

sin(t)j - cos(t)i

Step-by-step explanation:

Let's start with A:

Position vector = cos(t)i

Velocity vector = -sin(t)i (differentiating the position vector)

acceleration vector = -cos(t)i   (differentiating the velocity vector)

Then we go to B:

Position vector = sin(t)j

Velocity vector = cos(t)j

acceleration vector = -sin(t)j

Relative acceleration = -cos(t)i - (-sin(t)j) = sin(t)j - cos(t)i

answer

d. 10

explanation

the distance from point y to wx is the height of the triangle.

using pythagoras theorem,

group similar terms to get,

this implies that,

take positive square root of both sides to get,

this implies that,

b. 1.25 hr

step-by-step explanation:

if you have a constant speed of 60 mph, then you have a ratio between 60 mi. to 1 hr. dividing both sides by 60 we get the ratio

1 mi. to

then mult. both sides by 75 we get the ratio

75 mi. to

simplifying we get

75 mi. to

or 1.25 hr.