Question

Can someone please help me with this physics question? I'm desperate!

A track team is practicing for a 4 x 100 m relay race. The first runner, Linda, is running at a constant speed of 8.6 m/s. The next runner, Jenny, will be starting from rest at the 80 m mark. She has an acceleration of 1.0m/s^2. Ideally the two runners meet at the 100m mark to hand over the baton. At this point, Jenny is still accelerating.

a) How long does it take Jenny to run from the 80-m mark to the 100-m mark?

b)At what distance behind Jenny should Linda be when Jenny starts running? (Assume for simplicity that there is no distance between the two runners when the switch happens.)

c)What’s Jenny’s speed at the 100m mark?

Answer 1

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

Step-by-step 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, t, 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