Wilma i. ball walks at a constant speed of 5.93 m/s along a straight line from point a to point b and then back from b to a at a constant speed of 3.15 m/s.
(a) what is wilma's average speed over the entire trip?
(b) what is wilma's average velocity over the entire trip?

a) The average speed is 4.11 m/s

b) The average velocity is zero.

Explanation:

Hi there!

a) The average speed is calculated as the traveled distance over the time taken to cover that distance:

a.s = d/t

Where:

a.s = average speed.

d = traveled distance.

t = time.

From A to B Wilma travels at a constant speed of 5.93 m/s. So, we can write the equation of average speed as follows:

5.93 m/s = d / t1

Where t1 is the time taken to travel the distance from A to B (d)

From B to A Wilma travels at 3.15 m/s, so the equation of average speed could be written as follows:

3.15 m/s = d / t2

Where t2 is the time taken to travel from B to A (d)

If we solve both equations for the time, we can get the total time of the trip by adding the expressions we obtain:

t1 = d / 5.93 m/s

t2 = d / 3.15 m/s

The total time of the trip is the sum of both times:

t1 + t2 = d / 5.93 m/s + d / 3.15 m/s

t1 + t2 = (3.15 m/s d + 5.93 m/s d) / (3.15 m/s)(5.93 m/s)

t1 + t2 = 9.08 m/s d / (3.15 m/s)(5.93 m/s)

The total traveled distance is 2 d and the total time is t1 + t2, so the average speed for the entire trip will be:

a.s = 2 · d / (t1 + t2)

a.s = 2 · d · ((3.15 m/s)(5.93 m/s) / 9.08 m/s d)

a.s = 4.11 m/s

The average speed is 4.11 m/s

b) The average velocity is calculated as the displacement over the elapsed time. The displacement is calculated as follows:

displacement = initial position - final position

In this case:

initial position = final position ( because Wilma returns to the starting point, there was no net displacement).

Then:

a.v = D/t

Where:

a.v = average velocity.

D = displacement

t = time.

a.v = 0/ t

a.v = 0

The average velocity is zero.

laticia draws the diagram below to show earth’s magnetic field.

answer

the arrows to the right of earth should point up.

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