A person walks first at a constant speed of 2 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 4 m/s. What is the average speed (in m/s) over the entire trip?

The average speed is 2.67 m/s

Explanation:

Constant Speed Motion

The speed of a moving object can be calculated as:

Where

x = distance

t = time

Solving for t:

The average speed is calculated as:

Where xt and tt are the total distance and time respectively.

The person walks at a constant speed of v1=2 m/s for a distance x. They take a time t1 to travel calculated by:

Then, the person walks back at v2=4 m/s the same distance x. Thus, the second time is:

The total time taken for the entire travel is:

Simplifying and substituting:

The total distance is xt=2x, thus the average speed is:

Simplifying:

The average speed is 2.67 m/s

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the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s