A bicyclist on an old bike (combined mass: 92 kg) is rolling down (no pedaling or braking) a hill of height 120 m. Over the course of the 384 meters of downhill road, she encounters a constant friction force of 261 Newton. If her speed at the top of the hill is 9 m/s, what is her speed at the bottom of the hill

V = 48.49m/s

Explanation:

Given the following information:

Combined mass = 92kg

Hill's height = 120m

Course of ride = 384m

Frictional force = 261N

Initial speed (u) = 9m/s

Final speed (v) = ?

Since we are looking for her speed at the bottom,

Time = distance/speed = 384m/9m.s

Time = 42.67s

we use the equation

H = V²/2g ( equation for maximum heigh of trajectory)

Therefore, plugging the values we have

120 = V²/2×9.8

V² = 9.8×120×2

V = √2352

V = 48.49m/s

a person with a mass of 106kg has a weight of 1060n

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.