Question #19092

you walk away from an intersection at the rate of 1.1m/s for 60s. then you suddenly reverse course and run toward the intersection at the rate of 6.0m/s. formulate a d-t equation for each part of your trip.

Expert's answer

Let's define away from theintersection (i.e. origin) as positive, and towards the intersection as negative.

Then, walking away with a speed v1,

d = v1*t = (1.1 m/s) * t, where if t is in s, d will be given to you in m.

Moving so for T = 60 seconds, the object will reach the distance of X = v1*T =

1.1 m/s * 60 s = 66 m.

Then turning around and moving towards the origin with the speed v2, the distance will be defined as

d = X - v2*t = 66 m - (6.0 m/s) * t.

Answers:

d1 = v1*t = (1.1 m/s) * t;

d2 = X - v2*t = 66 m - (6.0 m/s) * t.

Then, walking away with a speed v1,

d = v1*t = (1.1 m/s) * t, where if t is in s, d will be given to you in m.

Moving so for T = 60 seconds, the object will reach the distance of X = v1*T =

1.1 m/s * 60 s = 66 m.

Then turning around and moving towards the origin with the speed v2, the distance will be defined as

d = X - v2*t = 66 m - (6.0 m/s) * t.

Answers:

d1 = v1*t = (1.1 m/s) * t;

d2 = X - v2*t = 66 m - (6.0 m/s) * t.

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