The position of an object moving along x-axis is given by x = a + bt² where a = 8.5m b = 2.5 ms-2 and t is measured in seconds. What is the average velocity between t = 2.0 s s and t = 4.0 s ?

The correct answer is 15 ms^-1.

Key Points

• The position of the object is given by the equation x = a + bt², where a = 8.5 m, b = 2.5 m/s^-2, and t is time in seconds.
• To find the average velocity, the formula used is: v_avg = (x_final - x_initial) / (t_final - t_initial).
• At t = 2.0 s, the position x_initial = 8.5 + 2.5(2.0)² = 18.5 m.
• At t = 4.0 s, the position x_final = 8.5 + 2.5(4.0)² = 48.5 m.
• The average velocity is calculated as v_avg = (48.5 - 18.5) / (4.0 - 2.0) = 30 / 2 = 15 ms^-1.

Additional Information

• Average Velocity: It is defined as the total displacement divided by the total time taken. It gives the overall rate of change of position over a time interval.
• Position Equation: The equation x = a + bt² represents a quadratic relation where acceleration is constant, and the motion is uniformly accelerated.
• Constant Acceleration: In the equation, the coefficient b (2.5 m/s²) represents half of the acceleration. The full acceleration can be derived as 2b = 5.0 m/s².
• Instantaneous Velocity: The instantaneous velocity at any time t can be found by differentiating x with respect to t: v = dx/dt = 2bt.
• Units of Measurement: Displacement is measured in meters (m), time in seconds (s), and velocity in meters per second (m/s).