A car travelling a `20 ms^(-1)` due north along the highway makes a right turn on to a side road that heads due east. It takes 50s for the car to complete the turn. At the end of 50 second, the car has a speed of `15 ms^(-1)` along the side road. Determine the magnitude of average acceleration over the 50 second interval.

Answer

To determine the magnitude of the average acceleration of the car over the 50-second interval, we can follow these steps:

### Step 1: Identify Initial and Final Velocities
The car's initial velocity (V_i) is given as:
- \( V_i = 20 \, \text{m/s} \) due north, which can be represented as a vector:
\[ V_i = 0 \hat{i} + 20 \hat{j} \, \text{m/s} \]

The car's final velocity (V_f) after the turn is:
- \( V_f = 15 \, \text{m/s} \) due east, represented as:
\[ V_f = 15 \hat{i} + 0 \hat{j} \, \text{m/s} \]

### Step 2: Calculate Change in Velocity
The change in velocity (ΔV) is calculated as:
\[ \Delta V = V_f - V_i \]
Substituting the values:
\[ \Delta V = (15 \hat{i} + 0 \hat{j}) - (0 \hat{i} + 20 \hat{j}) = 15 \hat{i} - 20 \hat{j} \, \text{m/s} \]

### Step 3: Determine Time Interval
The time interval (Δt) over which this change occurs is given as:
\[ \Delta t = 50 \, \text{s} \]

### Step 4: Calculate Average Acceleration
Average acceleration (a_avg) is given by the formula:
\[ a_{avg} = \frac{\Delta V}{\Delta t} \]
Substituting the change in velocity and time:
\[ a_{avg} = \frac{15 \hat{i} - 20 \hat{j}}{50} = \left(\frac{15}{50} \hat{i} - \frac{20}{50} \hat{j}\right) = 0.3 \hat{i} - 0.4 \hat{j} \, \text{m/s}^2 \]

### Step 5: Calculate Magnitude of Average Acceleration
The magnitude of the average acceleration is calculated using the Pythagorean theorem:
\[ |a_{avg}| = \sqrt{(0.3)^2 + (-0.4)^2} \]
Calculating the squares:
\[ |a_{avg}| = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \, \text{m/s}^2 \]

### Final Answer
The magnitude of the average acceleration over the 50-second interval is:
\[ \boxed{0.5 \, \text{m/s}^2} \]