Starting from rest a vehicle accelerates at the rate of \(2 \ m/s^2\) towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of \(4\sqrt{2} \ m/s^2\) for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is
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.
Similar Questions
Which one of the following equations related to the motion of an object is NOT correct? (Symbols carry their usual meanings)
Important Questions from Motion
The rate of change in the velocity of an object per unit time is referred as ________.
Which of the following is a correct equation of motion?
The acceleration of an object is said to be _______ when an object travels in a straight line and its velocity increases or decreases by an equal amounts in equal intervals of time.
What is the friction force employed between the two surfaces interacted in relative speed?
The rate of change of momentum of an object is
