An upright image which reduced in size 10 times occurred in a mirror. If the radius of curvature of the mirror is 2 m, bow far is the object from the mirror?
How to solve?​

p = -9 m

Explanation:

For this exercise we use the equation of the geometric optics constructor

where f is the focal length, p and q are the distance to the object and the image, respectively.

The mirrors the focal length is

         f = R / 2

         f = 2/2

         f = 1 m

the magnification is

         m =

indicates that the image was reduced h ’= h/10 implies that m = 1/10  

we write our system of equations

            p = -10q

            1/1 =

we substitute

           1 =

           1 = 1/p  (1 - 10)

           1 = -9 / p

           p = -9 m

good for her? ? where is the question?

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