An athlete speeds up uniformly from rest to a velocity of 10m/s after 2.0s.

Question

a) How far does the athlete travel in the first 1.0s?

b) How long does it take the athlete to travel 120m? (use a graphical method)

 

a) is 2.5m

b) is 13s

 

HOW?

Answer 1

(a)

 

d(t) = at²/2 + v₀t

 

v₀ = 0  [starts from rest]

 

a = Δv/Δt = (10 - 0)/2 m/s² = 5 m/s²

 

d(t) = (5/2)t²

 

d(1) = (5/2)(1)² m = 2.5 m

 

(b)

 

This forum is not set up for graphing.

 

t = total travel time

d₁ = distance traveled at constant acceleration of 5 m/s² for first 2 sec

d₂ = distance traveled at constant speed of 10 m/s for the next (t - 2) sec 

 

Set 120 m = d₁ + d₂ and solve for t.

 

120 = 2.5(2)² + 10(t - 2)

 

t - 2 = 11

 

t = 11 + 2 = 13 sec