The question asks for the average velocity of an object undergoing constant acceleration between specific times.
Given Information
- Initial velocity, \(v_0 = 0 \text{ m/s}\) (starts from rest)
- Acceleration, \(a = 2 \text{ m/s}^2\)
- Initial time, \(t_1 = 1 \text{ s}\)
- Final time, \(t_2 = 5 \text{ s}\)
Calculating Velocities at Specific Times
We use the kinematic equation for velocity under constant acceleration: \(v(t) = v_0 + at\).
- Velocity at \(t_1 = 1\) s: \(v_1 = 0 \text{ m/s} + (2 \text{ m/s}^2)(1 \text{ s}) = 2 \text{ m/s}\)
- Velocity at \(t_2 = 5\) s: \(v_2 = 0 \text{ m/s} + (2 \text{ m/s}^2)(5 \text{ s}) = 10 \text{ m/s}\)
Calculating Average Velocity
For motion with constant acceleration, the average velocity (\(v_{\text{avg}}\)) between two times can be calculated as the average of the initial and final velocities within that interval.
Formula: \(v_{\text{avg}} = \frac{v_1 + v_2}{2}\)
Calculation:
\( v_{\text{avg}} = \frac{2 \text{ m/s} + 10 \text{ m/s}}{2} \) \( v_{\text{avg}} = \frac{12 \text{ m/s}}{2} \) \( v_{\text{avg}} = 6 \text{ m/s} \)Final Answer
The average velocity between 1 s and 5 s is 6 m/s.
Similar Questions
- The area under the velocity–time curve between times \(t_1\) and \(t_2\) is equal to the ______ of the object during that interval of time.
- A train travels 45 m in 3 s. What is the speed of the train in km/h?
- A ball is dropped from a height and falls freely for 2 seconds. Its average speed during this time is: (Consider: acceleration due to gravity =\(10 \text{ m/s}^2\))
Important Questions from Speed and Velocity
Which of the following quantities specifies its speed with direction?
The basic unit of speed of an object is ______.
- Which of the following represents non-uniform linear velocity?
- Which of the following describes the nature of the average speed?
- The area under the velocity–time curve between times \(t_1\) and \(t_2\) is equal to the ______ of the object during that interval of time.
