B-1: Seth is doing his student driving with the "Give-Me-A-Brake” driving school
and is traveling down the interstate with a speed of 9.0 m/s. Mack is driving
his "18-wheeler" down the fast lane at 27.0 m/s when he notices Seth 30.0 m
ahead of him in the right lane. a) If Mack and Seth maintain their speeds, how
far must Mack travel before he catches up to Seth? b) How long will this take?

(a) t = 1.67 s

(b) s₂ = 45 m

Explanation:

Here, we use the formula:

s = vt

FOR Seth:

s₁ = v₁t₁

where,

s₁ = distance covered by Seth

v₁ = speed of Seth = 9 m/s

t₁ = time taken by Seth

FOR Mack:

s₂ = v₂t₂

where,

s₂ = distance covered by Mack

v₂ = speed of Mack = 27 m/s

t₂ = time taken by Mack

since, initially Mack is 30 m behind Seth. Therefore,

(a)

s₂ = s₁ + 30 m

using formulae:

v₂t₂ = v₁t₁ + 30 m

but, the time of catching is same for both (t₁ = t₂ = t)

v₂t = v₁t + 30 m

using values:

(27 m/s)t - (9 m/s)t = 30 m

t = (30 m)/(18 m/s)

t = 1.67 s

(b)

s₂ = v₂t

using values:

s₂ = (27 m/s)(1.67 s)

s₂ = 45 m

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.

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