Vehicle Motion Analysis: Calculating Net Displacement

This problem involves calculating the total displacement of a vehicle that undergoes motion in two distinct phases with different accelerations and directions. Displacement is a vector quantity, meaning it has both magnitude and direction. The net displacement is the overall change in position from the starting point to the final point.

Phase 1: Eastward Acceleration Calculation

The vehicle starts its journey from rest and accelerates towards the east. We can determine the distance covered during this phase using the kinematic equations of motion.

• Initial velocity (\(u_1\)): 0 m/s (since it starts from rest)
• Acceleration (\(a_1\)): \(2 \ m/s^2\) towards East
• Time duration (\(t_1\)): $10 \ s

We use the kinematic equation \(s = ut + \frac{1}{2}at^2\) to find the displacement (\(s_1\)) during this phase:

Calculation:

\(s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2\) \(s_1 = (0 \ m/s)(10 \ s) + \frac{1}{2} (2 \ m/s^2)(10 \ s)^2\) \(s_1 = 0 + \frac{1}{2} (2 \ m/s^2)(100 \ s^2)\) \(s_1 = 100 \ m\)

So, the displacement during the first phase is 100 m towards the East.

Phase 2: Southward Acceleration Calculation

After the first phase, the vehicle stops suddenly. This implies it comes to a complete halt before starting the next phase of motion. It then accelerates again towards the South.

• Initial velocity (\(u_2\)): 0 m/s (starts from rest after stopping)
• Acceleration (\(a_2\)): \(4\sqrt{2} \ m/s^2\) towards South
• Time duration (\(t_2\)): $10 \ s

Using the same kinematic equation \(s = ut + \frac{1}{2}at^2\), we find the displacement (\(s_2\)) during this second phase:

Calculation:

\(s_2 = u_2 t_2 + \frac{1}{2} a_2 t_2^2\) \(s_2 = (0 \ m/s)(10 \ s) + \frac{1}{2} (4\sqrt{2} \ m/s^2)(10 \ s)^2\) \(s_2 = 0 + \frac{1}{2} (4\sqrt{2} \ m/s^2)(100 \ s^2)\) \(s_2 = (2\sqrt{2})(100) \ m\) \(s_2 = 200\sqrt{2} \ m\)

The displacement during the second phase is \(200\sqrt{2}\) m towards the South.

Net Displacement: Vector Sum Calculation

The net displacement is the vector sum of the displacements from both phases. Since the first displacement (\(s_1\)) is towards the East and the second displacement (\(s_2\)) is towards the South, these two displacements are perpendicular to each other.

We can represent these displacements as vectors:

• \(\vec{s_1} = 100 \ m\) East
• \(\vec{s_2} = 200\sqrt{2} \ m\) South

To find the magnitude of the net displacement (\(\vec{S}\)), we can use the Pythagorean theorem, as the directions are perpendicular:

Calculation:

\(|\vec{S}| = \sqrt{s_1^2 + s_2^2}\) \(|\vec{S}| = \sqrt{(100 \ m)^2 + (200\sqrt{2} \ m)^2}\) \(|\vec{S}| = \sqrt{10000 \ m^2 + (40000 \times 2) \ m^2}\) \(|\vec{S}| = \sqrt{10000 \ m^2 + 80000 \ m^2}\) \(|\vec{S}| = \sqrt{90000 \ m^2}\) \(|\vec{S}| = 300 \ m\)

Therefore, the net displacement of the vehicle from its starting point is 300 m.