A runner moves along a straight track. For the first 4 seconds, she runs at a constant velocity of 5 m/s. For the next 4 seconds, she decelerates uniformly and comes to rest. What is the total distance she covers in 8 seconds?
30 meters
For straight-line movement without turning back, the distance covered equals the area under the velocity-time graph. The runner's journey has two stages, so the distance of each stage is found separately and the two are added.
Stage one is uniform velocity: distance = velocity × time = 5 m/s × 4 s = 20 m. On the velocity-time graph this is a rectangle of height 5 m/s and width 4 s.
Stage two is uniform deceleration from 5 m/s to 0 in 4 s. The graph is a triangle, so distance = \(\frac{1}{2} \times 4 \times 5\) = 10 m (equivalently, average velocity \(\frac{5 + 0}{2}\) = 2.5 m/s for 4 s).
Total distance in 8 seconds = 20 m + 10 m = 30 m.
Check: in stage two the deceleration is \(a = \frac{0 - 5}{4} = -1.25\) m/s2, and \(s = ut + \frac{1}{2}at^2 = 5 \times 4 - \frac{1}{2} \times 1.25 \times 16 = 20 - 10 = 10\) m, which agrees, so the total is again 20 + 10 = 30 m.
Hence, 30 meters.
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