Question

•• Al and Bert are jogging side-by-side on a trail in the woods at a speed of 0.75 m/s. Suddenly Al sees the end of the trail 35 m ahead and decides to speed up to reach it. He accelerates at a constant rate of 0.50 m/s2, while Bert continues on at a constant speed. (a) How long does it take Al to reach the end of the trail? (b) Once he reaches the end of the trail, he immediately turns around and heads back along the trail with a constant speed of 0.85 m/s. How long does it take him to meet up with Bert? (c) How far are they from the end of the trail when they meet?

Answer 1

a)

Consider the motion of Al :

v₀ = initial speed of Al = 0.75 m/s

d = distance to be traveled to reach the end of trail = 35 m

a = acceleration = 0.50 m/s²

t = time of travel

Using the equation

d = v₀ t + (0.5) a t²

inserting the values

35 = (0.75) t + (0.5) (0.50) t²

t = 10.4 sec

So Al reach the end of trail after 10.4 sec

b)

Consider the motion of Bert :

x = distance traveled by Bert when Al reach the end = ?

$v_{b}$ = speed of Bert = 0.75 m/s

t = time of travel = 10.4 sec

Distance traveled by Bert when Al reach the end is given as

x = $v_{b}$ t

x = (0.75) (10.4)

x = 7.8

x' = distance remaining to reach the end of trail for Bert = ?

distance remaining to reach the end of trail for Bert is given as

x' = d - x

x' = 35 - 7.8

x' = 27.2 m

$v_{rel}$ = relative velocity of Al with respect to bert = 0.85 + 0.75 = 1.6 m/s

t' = time to meetup

Time to meetup is given as

t' = x' /$v_{rel}$

t' = 27.2/1.6

t' = 17 sec

c)

L = length of the distance of the two from the end

$v_{AL}$ = speed of AL = 0.85 m/s

t' = time to meetup = 17 sec

length of the distance of the two from the end is given as

L = $v_{AL}$ t'

L = (0.85 (17)

L = 14.45 m