# Answer on Mechanics | Relativity Question for Sam

Question #11583

Barb is driving from Calgary to Edmonton at a speed of 120 km/h. Frank is travelling from Edmonton to Calgary at a speed of 112 km/h. Edmonton and Calgary are 298 km. How far from Calgary will they pass each other if they leave at the same time? How far from Calgary will they pass each other if Barb begins her journey 20 minutes before Frank?

Expert's answer

Let's assign the Calgary as a zero distance point. Then

120t = 298 - 112t ==> 120t + 112t = 298 ==> 232t = 298 ==> t = 149/116 ≈ 1.2845 h,

where t is the meeting time. So, they'll meet

120t = 120*149/116 = 4470/29 ≈ 154.1379 km from Calgary.

if Barb begins her journey 20 minutes (=1/3 hour) before Frank they'll meet after

120(t+1/3) = 298 - 112t ==> 120t + 40 = 298 - 112t ==> 232t = 258 ==> t = 129/116 ≈ 1.1121 h,

at the

120*129/116 = 3870/29 ≈ 133.4483 km

of the Calgary-Edmonton road.

120t = 298 - 112t ==> 120t + 112t = 298 ==> 232t = 298 ==> t = 149/116 ≈ 1.2845 h,

where t is the meeting time. So, they'll meet

120t = 120*149/116 = 4470/29 ≈ 154.1379 km from Calgary.

if Barb begins her journey 20 minutes (=1/3 hour) before Frank they'll meet after

120(t+1/3) = 298 - 112t ==> 120t + 40 = 298 - 112t ==> 232t = 258 ==> t = 129/116 ≈ 1.1121 h,

at the

120*129/116 = 3870/29 ≈ 133.4483 km

of the Calgary-Edmonton road.

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