Question

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?]

Answer 1

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

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

Answer 2

16.197 seconds

Explanation:

the time and the final place fot both cars is the same.

for the Boxster we have

X = Vboxster . T

being

X the distancance from the initial place of the Boxter to the meeting point

Vboxster the speed of the Boxster (26m/s)

T the time

and for the Scion

X = Xo + Vscion . T

Being

Xo the initial point of the Scion (115m)

X the distancance from the initial place of the Boxster to the meeting point

Vscion the speed of the Scion (18.9 m/s)

T the time

Xo + Vscion . T = Vboxster . T

Xo = Vboxster . T - Vscion . T

Xo = (Vboxster - Vscion) . T

Xo/(Vboxster - Vscion) = T

115m/ (26-18.9) m.s-1 = T

16.197 s = T