An object moves along a straight path for 5 s. Its initial velocity is 10 m s-1 and its final velocity is -2 m s-1. What is the magnitude of the average velocity?
4 m/s
Average velocity is total displacement divided by total time, not the simple average of the two speeds. Because the path is straight and the motion is uniformly accelerated, the displacement equals the average of the initial and final velocities multiplied by time: \(s = \frac{u + v}{2} \times t\).
Here \(u = 10\) m/s and \(v = -2\) m/s (the negative sign shows the object reversed direction). So \(s = \frac{10 + (-2)}{2} \times 5 = \frac{8}{2} \times 5 = 4 \times 5 = 20\) m.
Average velocity \(= \frac{\text{displacement}}{\text{time}} = \frac{20}{5} = 4\) m/s.
Hence, the magnitude of the average velocity is 4 m/s. Note that 6 m/s would be the plain average of the speeds 10 and 2, which is wrong because velocity is a vector and the directions are opposite; 8 m/s and 5 m/s do not follow from the displacement calculation.
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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?
- An object starts from rest at x = 0 m and t = 0 s. It moves with a constant acceleration of \(2 \text{ m/s}^2\) along the x-axis. What is its average velocity between time 1 s and 5 s?
