Question

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.2 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.6 m/s2 until he catches his friend.

A.) How much time does it take until he catches his friend (after his friend passes him)?

B.) How far has he traveled in this time?

C.) What is his speed when he catches up?

Answer 1

(a) First of all, after two seconds, the constant velocity bike will lead the bike at rest by a distance of:

d = st
= (3.2m/s)(2.0s)
= 6.4m

Call the accelerating bike, 1 and the constant velocity bike, 2. When 1 catches 2, it will have moved through a displacement of 6.4m + x. So the displacement of bike 1 is:

Δx₁ = v₀t + 0.5at²
6.4m + x = 0 + 0.5(2.6m/s²)t²
6.4m + x = (1.3m/s²)t² --------------->(1)

During this time, 2 goes through a displacement of:

Δx₂ = v₀t + 0.5at²
x = (3.2m/s)t + 0
x = (3.2m/s)t------------>(2)

By subtracting (2) from (1), we can eliminate x, and solve for t:

6.4m + x - x = (1.3m/s²)t² - (3.2m/s)t
6.4m/s = (1.3m/s²)t² - (3.2m/s)t

this is a quadratic equation, in standard form:

(1.3m/s²)t² - (3.2m/s)t - 6.4m/s = 0

t = 3.77s or t = -1.30s

The minus root is meaningless, so the time for 1 to catch 2 is 3.77s.

(b) In this time, 1 has moved through a displacement of:

Δx₁ = v₀t + 0.5at²
= 0 + 0.5(2.6m/s²)(3.77s)²
= 18.477m

(c) His speed after this time is:

v = v₀ + at
= 0 + (2.6m/s²)(3.77s)
= 9.802m/s