Answer:
a) 2·√10 seconds
b) Linda should be approximately 30.6 meters
c) Jenny's speed at the 100-m mark is approximately 6.325 m/s
Explanation:
The speed with which Linda is running = 8.6 m/s
The point Jenny starts = The 80-m mark
The acceleration of Jenny = 1.0 m/s²
a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows
Δs = u·t + (1/2)·a·t²
Δs = Distance = 100-m - 80-m = 20-m
u = The initial velocity of Jenny = 0
a = Jenny's acceleration = 1.0 m/s²
∴ 20 = 0×t + (1/2) × 1 × t² = t²/2
20 = t²/2
t = √(20 × 2) = 2·√10
The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds
b) The distance Linda runs in t = 2·√10 seconds, d = v × t
Given that Linda's velocity, v = 8.6 m/s, we have;
d = 8.0 × 2·√10 = 16·√10
The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters
Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running
c) Jenny's speed at the 100 m mark is given as follows;
v = u + a·t
t = 2·√10 seconds, a = 1.0 m/s², u = 0
∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325
Jenny's speed at the 100-m mark ≈ 6.325 m/s
