Answer:
(a) 5 s
(b) 20 m/s
(c) 200 m
Explanation:
Part a
The average velocity (vₐ) is the total displacement (s) divided by the total time (T). On a velocity-time graph, the total displacement is equal to the area under the curve. In this case, the region under the curve is a trapezoid. The area of a trapezoid is the average of the parallel sides (a and b) times the perpendicular height (h). So the total displacement of the particle is:
s = ½ (a + b) h
s = ½ (20 − 2t + 20) v
s = (20 − t) v
The acceleration is the slope of the line. Given that the acceleration is 4 m/s², the maximum velocity is v = 4t. Substituting, the displacement is:
s = (20 − t) 4t
s = 80t − 4t²
Given that the average velocity is 15 m/s:
vₐ = s / T
15 = (80t − 4t²) / 20
300 = 80t − 4t²
4t² − 80t + 300 = 0
t² − 20t + 75 = 0
(t − 5) (t − 15) = 0
t = 5 or 15
Since 20 − 2t must be positive, t can't be 15. Therefore, t = 5 seconds.
Part b
The maximum velocity is:
v = 4t
v = 20 m/s
Part c
The distance traveled with uniform velocity is:
d = v (20 − 2t)
d = 200 m
