Help will be appreciated.

In a large department store, a shopper is standing on the "up" escalator, which is travelling at an angle of 40 degree above the horizontal and at a speed of 0.75 m/s. He passes his daughter, who is standing on the identical, adjacent "down" escalator. Find the velocity of the shopper relative to his daughter in unit vector notation.

0.96j m/s

Well, it seems the shopper and his daughter are going in opposite directions. That's quite an escalator situation they've got going on!

To find the velocity of the shopper relative to his daughter, we need to consider their velocities separately.

Let's break it down step by step. First, we have the shopper's velocity:

The shopper is moving up the escalator at an angle of 40 degrees above the horizontal. Given that the speed of the escalator is 0.75 m/s, we can break it down into its horizontal and vertical components.

The horizontal component of the shopper's velocity is Vx = Vcosθ, where V is the speed of the shopper and θ is the angle. So, Vx = 0.75 m/s * cos(40°).

The vertical component of the shopper's velocity is Vy = Vsinθ, where V is the speed of the shopper and θ is the angle. So, Vy = 0.75 m/s * sin(40°).

Next, let's look at the daughter's velocity:

Since the daughter is standing on the "down" escalator, her velocity will be the negative of the velocity of the "up" escalator. So, the daughter's velocity will be -0.75 m/s.

Now, to find the velocity of the shopper relative to his daughter, we subtract the daughter's velocity from the shopper's velocity:

Vrelative = (Vx - 0) i + (Vy - (-0.75)) j.

Simplifying it:

Vrelative = (0.75 m/s * cos(40°)) i + (0.75 m/s * sin(40°) + 0.75 m/s) j.

So, the velocity of the shopper relative to his daughter is approximately (0.452 i + 1.322 j) m/s.

I hope that helps! And remember, at least they're not both stuck on the "down" escalator. Keep that humor switched on!

To find the velocity of the shopper relative to his daughter in unit vector notation, we can break it down into horizontal and vertical components.

Let's assume that the right direction is the positive x-direction and upward is the positive y-direction.

The velocity of the shopper can be expressed as:
V_shopper = V_x_shopper * i + V_y_shopper * j

Given that the shopper is standing on the "up" escalator, the velocity can be determined by the angle and speed of the escalator:
V_x_shopper = 0.75 m/s * cos(40°) = 0.573 m/s (positive because it is in the same direction as the rightward x-axis)
V_y_shopper = 0.75 m/s * sin(40°) = 0.483 m/s (positive because it is in the same direction as the upward y-axis)

Now let's determine the velocity of the daughter. Since she is on the "down" escalator, her velocity would be given as:
V_x_daughter = -0.75 m/s * cos(40°) = -0.573 m/s (negative because it is in the opposite direction of the rightward x-axis)
V_y_daughter = -0.75 m/s * sin(40°) = -0.483 m/s (negative because it is in the opposite direction of the upward y-axis)

To find the velocity of the shopper relative to his daughter, we simply subtract the velocity of the daughter from the velocity of the shopper:
V_relative = (V_x_shopper - V_x_daughter) * i + (V_y_shopper - V_y_daughter) * j

Substituting the values we calculated earlier:
V_relative = (0.573 m/s - (-0.573 m/s)) * i + (0.483 m/s - (-0.483 m/s)) * j
V_relative = (1.146 m/s) * i + (0.966 m/s) * j

Therefore, the velocity of the shopper relative to his daughter in unit vector notation is (1.146 m/s)i + (0.966 m/s)j.

To find the velocity of the shopper relative to his daughter, we need to consider the vectors involved in this scenario.

Let's break down the problem:

1. The shopper is standing on the "up" escalator, which is traveling at an angle of 40 degrees above the horizontal and at a speed of 0.75 m/s.
2. The daughter is standing on the identical, adjacent "down" escalator.

To find the velocity of the shopper relative to his daughter, we need to subtract the velocity of the daughter from the velocity of the shopper.

Step-by-step solution:

1. Convert the given speed and angle of the "up" escalator into components.
- The vertical component of the velocity is 0.75 m/s (upward since the escalator is going up).
- The horizontal component of the velocity is 0.75 m/s * cos(40°) (to the right since the angle is above the horizontal).
- Calculate the horizontal component: 0.75 m/s * cos(40°) ≈ 0.57 m/s.
- Calculate the vertical component: 0.75 m/s * sin(40°) ≈ 0.48 m/s upward.

2. Since the "down" escalator is identical to the "up" escalator, the speed and angle are the same, but the direction is different.

3. Convert the speed and angle of the "down" escalator into components.
- The vertical component of the velocity is 0.75 m/s (downward since the escalator is going down).
- The horizontal component of the velocity is 0.75 m/s * cos(40°) (to the right since the angle is above the horizontal).
- Calculate the horizontal component: 0.75 m/s * cos(40°) ≈ 0.57 m/s.
- Calculate the vertical component: -0.75 m/s * sin(40°) ≈ -0.48 m/s downward (negative since the direction is downward).

4. Subtract the components of the daughter's velocity from the components of the shopper's velocity to get the relative velocity.
- Horizontal component: 0.57 m/s - 0.57 m/s = 0 m/s (since they have the same horizontal component).
- Vertical component: 0.48 m/s - (-0.48 m/s) = 0.96 m/s upward (subtracting negative values is the same as adding).

5. Write the relative velocity vector in unit vector notation.
- The unit vector notation represents a vector as <x-component, y-component>. In our case, since the horizontal component is 0 and the vertical component is 0.96 m/s, the relative velocity vector is <0, 0.96> m/s.

Therefore, the velocity of the shopper relative to his daughter, in unit vector notation, is <0, 0.96> m/s.