Question

1.
A cyclist is traveling at 12.0 m/s when he applies the brakes and begins to slow. After
4.5 s the cyclist has slowed to 3.0 m/s. What was the cyclist's (a) acceleration (assume it
was constant), and (b) the distance traveled in this time?

Answer 1

ans: a) -2.0m/s^2
b) 33.75m

-List our knows and unknowns:

vi=12.0m/s
vf=3.0m/s
t=4.5s
a=?
d=?

-a) solve for a. let’s choose a formula, avoiding d, as we do not want two unknown variables when solving for one. Here’s a good one:

vf=vi+at

-Plug in your values:

3.0=12.0+a(4.5)
-9.0=4.5a
-9/4.5=a
a=-2.0m/s^2

-b) solve for d. let’s choose a formula. Here’s a good one:

d=vit + 1/2at^2

-plug in your values:

d=(12.0)(4.5)+1/2(-2.0)(4.5^2)
d=(54)+(-20.25)
d=33.75m