Question

Suppose that a particle moves on a coordinate line so that its velocity at time $t$ is $v(t) = t^2 - 2t \text{ m/s}$ , then the displacement of the particle during the time interval $0 \leq t \leq 3$ is given by}

• A. 0 m
• B. 1 m
• C. 1/2 m
• D. 3 m

Hint:

Displacement measures net change in position (which can be positive, negative, or zero), whereas total distance traveled is the integral of the absolute value of velocity, $\int |v(t)| \, dt$, which is always non-negative.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Displacement is defined as the change in position of an object and is calculated as the definite integral of its velocity function over a given time interval.
Key Formula or Approach:
The formula for displacement over the time interval $[t_1, t_2]$ is: \[ \text{Displacement} = \int_{t_1}^{t_2} v(t) \, dt \]

Step 2: Detailed Explanation:
We are given the velocity function: \[ v(t) = t^2 - 2t \] And the time interval is $[0, 3]$.
Calculate the displacement by integrating the velocity function: \[ \text{Displacement} = \int_0^3 (t^2 - 2t) \, dt \] \[ = \left[ \frac{t^3}{3} - t^2 \right]_0^3 \]
Evaluate this expression at the upper limit ($t = 3$): \[ \left( \frac{3^3}{3} - 3^2 \right) = \left( \frac{27}{3} - 9 \right) = 9 - 9 = 0 \]
Evaluate at the lower limit ($t = 0$): \[ \left( \frac{0^3}{3} - 0^2 \right) = 0 \]
Subtracting the lower limit value from the upper limit value: \[ \text{Displacement} = 0 - 0 = 0\text{ m} \]
This means that although the particle moved during the interval, it returned to its exact starting position at $t = 3$.

Step 3: Final Answer:
The displacement of the particle is 0 m.