A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway and 4.0 s later,

its instantaneous velocity is 16 m/s, 45o north of west. What is the magnitude of the average acceleration
of the car during the five-second interval?

To find the magnitude of the average acceleration of the car during the five-second interval, we can use the formula:

Average acceleration = (change in velocity) / (time interval)

First, let's find the change in velocity:

The initial velocity of the car is 24 m/s, west.
The final velocity of the car is 16 m/s, 45° north of west.

To find the change in velocity, we need to subtract the initial velocity from the final velocity.

Change in x-component of velocity = 16 m/s * cos(45°) - 24 m/s * cos(0°)
Change in x-component of velocity = 11.31 m/s - 24 m/s = -12.69 m/s

Change in y-component of velocity = 16 m/s * sin(45°) - 24 m/s * sin(0°)
Change in y-component of velocity = 11.31 m/s - 0 m/s = 11.31 m/s

Now, we can find the magnitude of the change in velocity:

Magnitude of the change in velocity = sqrt((Change in x-component of velocity)^2 + (Change in y-component of velocity)^2)
Magnitude of the change in velocity = sqrt((-12.69 m/s)^2 + (11.31 m/s)^2)
Magnitude of the change in velocity = sqrt(161.0161 m^2/s^2 + 127.5361 m^2/s^2)
Magnitude of the change in velocity = sqrt(288.5522 m^2/s^2)
Magnitude of the change in velocity ≈ 16.99 m/s

Now, let's find the time interval:

The time interval is given as 5.0 seconds.

Finally, we can calculate the average acceleration:

Average acceleration = (magnitude of change in velocity) / (time interval)
Average acceleration = 16.99 m/s / 5.0 s
Average acceleration ≈ 3.398 m/s^2

Therefore, the magnitude of the average acceleration of the car during the five-second interval is approximately 3.398 m/s^2.

To find the magnitude of the average acceleration of the car during the five-second interval, we need to use the formula for average acceleration.

Average acceleration (a) is defined as the change in velocity (Δv) divided by the change in time (Δt):

a = Δv / Δt

First, we need to find the change in velocity (Δv):

The car's initial velocity is 24 m/s, west.
The car's final velocity is 16 m/s, 45° north of west.

To find the change in velocity, we need to calculate the vector difference between the final velocity and the initial velocity. We'll break down the final velocity into its north and west components.

The north component of the final velocity is given by: 16 m/s * sin(45°)
The west component of the final velocity is given by: 16 m/s * cos(45°)

Now, we can calculate the change in velocity:

Δv = (final north component - initial north component) + (final west component - initial west component)

Δv = [(16 m/s * sin(45°)) - 0] + [0 - (-24 m/s)] (since the initial north and west components are both 0)

Δv = (16 * sqrt(2) - 0) + (0 - (-24))

Δv = (16 * sqrt(2)) + 24

Next, we need to find the change in time (Δt). The problem tells us that the car exits the highway 4.0 s later. So, Δt = 5.0 s - 4.0 s = 1.0 s.

Now, we can calculate the average acceleration:

a = Δv / Δt

a = [(16 * sqrt(2)) + 24] / 1.0

a = 16 * sqrt(2) + 24

Therefore, the magnitude of the average acceleration of the car during the five-second interval is 16 * sqrt(2) + 24.

Is it a really five second interval? You only provide information for a four second interval. I'd need to know what the car did during the last second.

Anyway, calculate the vector difference in the car velocities (final minus initial) , and divide the result by the time interval, whatever it is.