Carefully analyze each problem.
Provide the necessary formula and solutions to support your
answer. Follow the GAFSA Rule: Given, Ask, Formula, Solution, and Answer.
Note: Answer must be in sentence form.
1. During a freestyle competition, a swimmer performs the crawl stroke in a pool 50.0 m long, as
shown in Figure. She swims a length at racing speed, taking 24.0 s to cover the length of the
pool. She then takes twice that time to swim casually back to her starting point. Find (a) her
average velocity for each length and (b) her average velocity for the entire swim.
Given: x=50.0 m
t=24.0 s
Asked: v for each length=?
v for the entire swim=?
x f −x i
Formula: v=
t f −t i
Solution: (a.1) average velocity for first length
x i=0 ; x f =50.0 m; t i=0 ; t f =24.0 s
x f −x i 50.0 m−0 50.0 m
v= = = =2.08 m/s
t f −t i 24.0 s−0 24.0 s
(a.1) average velocity for return trip
x i=50.0 m; x f =0 ; t i=24.0 s ; t f =24.0 s+ 48.0 s=72.0 s
x f −x i 0−50.0 m −50.0 m
v= = = =−1.04 m/s
t f −t i 72.0 s−24.0 s 48.0 s
(b) average velocity for the entire swim
x f −x i 50.0 m−50.0 m 0
v= = = =0
t f −t i 72.0 s−24.0 s 48.0 s
Answer: The average velocity for each length are 2.08 m/s (first length) and -1.04 m/s (return trip
length). On the other hand, the average velocity for the entire swim is 0 because there was no net
change in position. Furthermore, the total displacement for a round trip is always 0.
2. A motorist traveling at a constant velocity of 15m/s passes a school-crossing corner where the
speed limit is 10 m/s (about 22 mi/h). A police officer on a motorcycle stopped at the corner
starts off in pursuit with constant acceleration of 3.0 m/s 2. (a) How much time elapses before
the officer catches up with the car? (b) What is the officer’s speed at that point? (c) What is the
total distance the officer has traveled at that point?
2
Given: Police: x P 0 =0 ; v P 0=0 ; a P=3.0 m/s
Car: x C 0=0 ; v C 0=15 m/s ; aC =0
Asked: x=?
v=?
t=?
1 2
Formula: x=x 0 + v 0 t + a t
2
v P=v 0 +a P t
1
x= a P t
2
1 2 1 2 2
Solution: Police x: x P =x P 0 + v P 0 t+ a P t =0+0 t + ( 3.0 ) t =1.5 t
2 2
1 2 1 2
Car x: x C =x C 0 +v C 0 t+ aC t =0+ 15t + ( 0 ) t =15 t
2 2
(a) time before the officer catches up with the car
x P =x C
2 2 15 1.5 t
15 t=1.5 t ; 15 t−1.5t =0 ; t (15−1.5 t )=0 ; 15−1.5 t=0 ; = =10
1.5 1.5
t=10 s
(b) officer’s speed at that point
v P=v 0 +a P t=0+ ( 3 m/s 2 ) ( 10 s )=30 m/s
(c) total distance the officer has traveled at that point
1 1
x= a P t= ( 3 m/s ) ( 10 s ) =150 m
2 2
2 2
Answer: The officer catches up with the car after 10 s of pursuit. He traveled a total of 150 m at
30 m/s at that point.
