Two cars, a Porsche Boxster and a Toyota Scion XB, are traveling in the same direction, although the Boxster is 115.0 m behind the Scion. The speed of the Boxster is 26.0 m/s and the speed of the Scion is 18.9 m/s. How much time does it take for the Boxster to catch the Scion? [Hint: What must be true about the displacement of the two cars when they meet?]

a) t = 16.2s : time it takes for the Boxster to catch the Scion

b) The displacement of the Boxster (db) is 115 m more than the displacement of the Scion (ds)

db = ds+115

Step-by-step explanation:

Conceptual analysis

We apply the formula for constant speed movement:

v = d/t Formula (1)

v = speed in m/s

d: distance in m

t: time in s

Problem development

The time of Scion (ts) is equal time of Boxster (tb)

t (s) = tb=t

The displacement of the Boxster is 115 m more than the displacement of the Scion

db = ds+115

we apply formula (1) car kinematics:

Scion kinematics

18.9=ds/t

t = ds / 18.9 Equation (1)

Boxster kinematics

26=db/t

26 = (ds+115) / t

t = (ds+115) / 26 Equation (2)

Equation (1) = Equation (2)

ds / 18.9 = (ds+115) / 26

18.9 (ds+115) = 26 ds

18.9ds+18.9*115=26 ds

2173.5 = 26 ds-18.9ds

2173.5=7.1ds

ds = 2173.5:7.1

ds=306.12m

We replace ds=306.12m in the equation (1)

t = 306.12:18.9

t = 16.2s