a car travels along a highway with a velocity of 24m/s, west. then exits the highway and 4s later, its instantaneous velocity is 16m/s, 45degrees north of west. Calculate the magnitude of the average accelaration of the car during the 5s interval?

i = unit vector x (east)

j = unit vector y (north)
Vo = -24 i + 0 j
Vf = -16/sqrt2 i + 16/sqrt2 j
change in velocity =(24-16/sqrt2)i + 16/sqrt2 j
dV/dt = A = (1/5)[(24-16/sqrt2)i + 16/sqrt2 j ]

To get the magnitude of A take the square root of the sum of the squares.

To calculate the magnitude of the average acceleration of the car during the 5-second interval, we need to find the change in velocity and divide it by the time interval.

Step 1: Find the initial velocity (v₁):
Given that the car travels along the highway with a velocity of 24 m/s, westward, the initial velocity (v₁) is -24 m/s since it is moving in the opposite direction of the positive x-axis.

Step 2: Find the final velocity (v₂):
4 seconds later, the car's instantaneous velocity is 16 m/s, 45 degrees north of west. We need to calculate the x and y components of this velocity.
The x-component of the velocity is given by 16 m/s * cos(45) = 11.31 m/s, westward (negative because it is in the opposite direction of the positive x-axis).
The y-component of the velocity is given by 16 m/s * sin(45) = 11.31 m/s, northward.

So the final velocity (v₂) is (11.31 m/s, 11.31 m/s), west-northwestward.

Step 3: Calculate the change in velocity (Δv):
The change in velocity is given by the final velocity (v₂) minus the initial velocity (v₁).
Δv = v₂ - v₁ = (11.31 m/s, 11.31 m/s) - (-24 m/s, 0) = (35.31 m/s, 11.31 m/s)

Step 4: Calculate the average acceleration (a):
The average acceleration is the change in velocity (Δv) divided by the time interval (t).
a = Δv / t = (35.31 m/s, 11.31 m/s) / 5s = (7.062 m/s², 2.262 m/s²)

Step 5: Calculate the magnitude of the average acceleration (|a|):
The magnitude of the average acceleration can be found using the Pythagorean theorem:
|a| = √(a_x² + a_y²) = √((7.062 m/s²)² + (2.262 m/s²)²) = √(49.733 m²/s⁴ + 5.122 m²/s⁴) ≈ √54.855 m²/s⁴ ≈ 7.408 m/s²

Therefore, the magnitude of the average acceleration of the car during the 5-second interval is approximately 7.408 m/s².

To calculate the magnitude of the average acceleration of the car during the 5-second interval, we need to find the change in velocity and divide it by the time taken. Here's how you can solve it step by step:

1. Determine the initial velocity (v₁) and final velocity (v₂):
- The initial velocity of the car is given as 24 m/s west.
- The final velocity of the car is given as 16 m/s, 45 degrees north of west.

2. Calculate the change in velocity (Δv):
- To find the change in velocity, we subtract the initial velocity from the final velocity.
Δv = v₂ - v₁

3. Convert the final velocity's direction to west component:
- Since the final velocity is given as 16 m/s, 45 degrees north of west, we need to find the west component.
- Using trigonometry, we can determine that the west component is v₂ * cos(45°).

4. Calculate the change in velocity in both magnitude and direction:
- To find the change in velocity magnitude, take the absolute value of the west component of the change in velocity.
|Δv| = |v₂west| - |v₁west|

- To find the direction of the change in velocity, use the angle formed between the west direction and the final velocity.
θ = atan2(v₂north, v₂west) - atan2(v₁north, v₁west)

5. Calculate the average acceleration (a_avg):
- The average acceleration is given by the change in velocity divided by the time interval.
a_avg = Δv / t

Substitute the given values into the equations to find the magnitude of the average acceleration during the 5-second interval.