Questions & Answers: "Motion and Kinematics"

The radius of the circular track is $R = 70 \text{ m}$. For half a revolution:

1. Distance covered: This is the actual path length traveled by the car, which is half the circumference of the circle. Distance = $\frac{1}{2} \times (2 \pi R) = \pi R = \frac{22}{7} \times 70 = 22 \times 10 = 220 \text{ m}$.
2. Displacement: This is the shortest straight-line distance from the initial position (point A) to the final position (point B). For half a revolution on a circular path, the final position B is diametrically opposite to the initial position A. Therefore, the displacement is equal to the diameter of the circle. Displacement = $2R = 2 \times 70 = 140 \text{ m}$. Thus, the distance covered is 220 m and the displacement is 140 m.

The total displacement of the object can be calculated by finding the area under its velocity-time graph. We can divide the motion into three segments:

1. Segment 1 (0 to 4 seconds): The object accelerates uniformly from rest (0 m/s) to 20 m/s. This forms a triangle on the velocity-time graph. Displacement $s_1 = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \text{ s} \times 20 \text{ m/s} = 40 \text{ m}$.
2. Segment 2 (4 to 6 seconds): The object moves at a constant velocity of 20 m/s for 2 seconds. This forms a rectangle on the velocity-time graph. Displacement $s_2 = \text{Area of rectangle} = \text{base} \times \text{height} = 2 \text{ s} \times 20 \text{ m/s} = 40 \text{ m}$.
3. Segment 3 (6 to 8 seconds): The object decelerates uniformly from 20 m/s to rest (0 m/s) in 2 seconds. This forms another triangle on the velocity-time graph. Displacement $s_3 = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \text{ s} \times 20 \text{ m/s} = 20 \text{ m}$. The total displacement is the sum of the displacements in each segment: Total Displacement = $s_1 + s_2 + s_3 = 40 \text{ m} + 40 \text{ m} + 20 \text{ m} = 100 \text{ m}$.

We can break the car's motion into three phases:

Phase 1: Uniform Acceleration Initial velocity ($u$) = 0 m/s (starts from rest) Acceleration ($a$) = $2 \text{ m/s}^2$ Time ($t$) = 5 s Final velocity ($v_1$) after this phase: $v_1 = u + at = 0 + (2 \text{ m/s}^2)(5 \text{ s}) = 10 \text{ m/s}$. Distance covered ($s_1$): $s_1 = ut + \frac{1}{2}at^2 = (0)(5) + \frac{1}{2}(2 \text{ m/s}^2)(5 \text{ s})^2 = 0 + 25 \text{ m} = 25 \text{ m}$.

Phase 2: Constant Velocity Velocity ($v$) = $10 \text{ m/s}$ (the final velocity from Phase 1) Time ($t$) = 10 s Distance covered ($s_2$): $s_2 = v \times t = (10 \text{ m/s})(10 \text{ s}) = 100 \text{ m}$.

Phase 3: Uniform Deceleration Initial velocity ($u$) = $10 \text{ m/s}$ (the constant velocity from Phase 2) Final velocity ($v$) = 0 m/s (comes to rest) Acceleration ($a$) = $-4 \text{ m/s}^2$ (deceleration) Distance covered ($s_3$): Using the equation $v^2 = u^2 + 2as$ $0^2 = (10 \text{ m/s})^2 + 2(-4 \text{ m/s}^2)s_3$ $0 = 100 - 8s_3$ $8s_3 = 100 \text{ m}$ $s_3 = \frac{100}{8} \text{ m} = 12.5 \text{ m}$.

Total distance covered by the car: Total Distance = $s_1 + s_2 + s_3 = 25 \text{ m} + 100 \text{ m} + 12.5 \text{ m} = 137.5 \text{ m}$.