Both Cars Start at the Same Point Where Xinitial = 0

Physics Page 50 # 8

Two cars travel in the same direction along a straight highway, one at a constant speed of 55 miles/hour and the other at 70 miles/hour. (a) Assuming that they start at the same location at the same time, how much sooner does the faster car arrive at a destination 10 miles away? (b) How far must the faster car travel before it has a 15 minute lead on the slower car?

Here I assumed that the least accurate value was to 2 significant figures (SF) so we must round our answer to 2 SF and use 3 SF for calculations.

Both cars start at the same point where xinitial = 0

Vslower = 55 mi/h and Vfaster = 70 mi/h

Destination = Δx = 10 mi

Distance formula: Distance = (Rate)(Time) or Time = (Distance)/(Rate)

(a) Determine the time needed for each car to go 10 miles.

Solve the distance formula for time

tslowercar = Time = (Distance)/(Rate)

tslowercar = (10 mi)/(55 mi/h) = (10h)/(55)

tslowercar = (2/11)h

tfaster = (10 mi)/(70 mi/h) = (10h)/(70)

tfaster = (1/7)h

Time between two cars when faster car gets 10 miles away = Δt

Δt = tslowercar - tfaster = (2/11)h – (1/7)h

Δt = (2/11)(7/7)h – (1/7)(11/11)h

Δt = (14/77)h – (11/77)h = (3/77)h

Δt = 0.038961938 h (Round to 3 SF)

Δt = 0.0390 h or Δt = (0.0390 h)(60 min/1 h) = 2.34 min

(b) Find distance for faster car to be 15 min ahead of slower car.

Vrelative = Vfaster – Vslower = 70 mi/h – 55 mi/h = 15 mi/h

Δt = Δx/ Vrelative = (Vslower)(time)/ Vrelative

Δt = (55 mi/h)(15 min)/(15 mi/h) = 55 min

Distance = (Rate)(time)

Δxfaster = (70 mi/h)(55 min)(1 h/60 min)
Δxfaster = (70 mi/h)(11 h)/(12) = (35 mi)(11)/(6)

Δxfaster = 64.166666667 (Round to 2 significant figures)

Δxfaster = 64 miles