A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a constant eastward acceleration of 0.75 m/s 2 over a distance of 20.0 m. Immediately after this 20.0 m distance, the ball is kicked northward in a manner that does not change its speed at that point, just its direction. The ball is allowed to come to a complete stop. How far north does the ball travel before stopping if the acceleration after this kick is 1.15 m/s 2

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : = 0.75 m/s², = 20 m, = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

² = (10)² + (2 × 0.75 × 20)

² = 100 + 30

² = 130

= √130

= 11.4 m/s

for the Second Course:

=  11.4 m/s,   = -1.15 m/s²,   = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×   )

0 = 129.96 - 2.3

2.3  = 129.96

= 129.96 / 2.3

= 56.5 m

so;

|d| = √( ² + ² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

wavelength variability is the right one

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)