In one day, a 70 kgkg mountain climber ascends from the 1500 mm level on a vertical cliff to the top at 2400 mm . The next day, she descends from the top to the base of the cliff, which is at an elevation of 1190 mm .

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

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..