What is the purpose of a bicycle to increase friction and what is to reduce friction What parts?

What is the purpose of a bicycle to increase friction and what is to reduce friction What parts?

Increase: tire, brake, handrail, pedal
Reduction: where there are ball bearings (such as wheels), the tire is round,

There are two cars a and B on the straight highway. A starts from a standstill with an acceleration of 0.5m/s2, and B makes a uniform motion in the same direction at a speed of 5m / s 200m in front of A. Q: (1) When will a catch up with B? How fast does a catch up with B? How far is a from the starting point? (2) When is the maximum distance between a and B in the process of catching up? What is the distance?

(1) When a catches up with B, their displacement difference is x0 = 200m,   X a = x0 + X B. suppose a catches up with B after time t, then x a = 12a a T2, x B = v b T. according to the catch-up conditions, 12a T2 = v b t + 200, the solution is t = 40   S or T = - 20   S (rounded). At this time, the speed of a V A = a a

There are two cars a and B on the straight highway. A starts from a standstill with an acceleration of 0.5m/s2, and B makes a uniform motion in the same direction at a speed of 5m / s 200m in front of A. Q: (1) When will a catch up with B? How fast does a catch up with B? How far is a from the starting point? (2) When is the maximum distance between a and B in the process of catching up? What is the distance?

(1) When a catches up with B, their displacement difference is x0 = 200m,
  X a = x0 + x B,
Let a catch up with B in time t, then x a = 1
2A a T2, x B = v b t
According to the requirements and conditions, there are 1
2A a T2 = v b t + 200,
The solution is t = 40   S or T = - 20   S (rounded off)
At this time, the speed of a V A = a T = 0.5 × forty   m/s=20   m/s，
Displacement of a from the starting point x a = 1
2A a T2 = 1
two × zero point five × four hundred and two   m=400   m．
(2) In the process of catching up, when the speed of a is less than that of B, the distance between a and B is still increasing, but when the speed of a is greater than that of B, the distance between a and b decreases. When the two speeds are equal, the distance between a and B reaches the maximum
By a a T = v b,
T = 10   s，
That is, a is at 10   The distance from the end of s to B is the largest
Xmax = x0 + V B T-1
2A a T2 = (200 + 5) × 10-1
two × zero point five × 102）   m=225   m．
A: (1) when a catches up with B in 40s, the speed when a catches up with B is 20   M / s, at this time, a is 400 away from the starting point   m．
(2) In the process of catching up, 10 between a and B   S has the maximum distance, which is 225   m．

There are two cars a and B on the straight highway. Car a starts from a standstill with the acceleration of a = 0.5m / S ^ 2, and car B starts at VO = 5m at XO = 200m in front of Party A/ Q: 1. When will a catch up with B? How fast did a catch up with B? How far is a from the starting point? 2. When is the maximum distance between a and B in the process of catching up? What is the distance?

1, s a = v0t + 1 / 2at ^ 2, V0 = 0. Set the time t after a catches up with B, S B + 200 = s a, 5T + 200 = 0.5 * T ^ 2, t = 40s,
At this time, V A = 0.5 * 40 = 20m / s, s a = 400m, so after 40 seconds, a catches up with B. at this time, the speed is 20m / s, and a is 400m from the starting point
2. When the speed of Party A and Party B is the same, the displacement is the largest, V A = 0.5 * t = 5m / s, t = 10s, and the distance L = s B + 200-s a = 5 * 10 + 200-0.25 * 100 = 125m, so the distance between Party A and Party B is the largest, 125m at 10 seconds

There are two cars a and B on the straight highway. A starts from a standstill with an acceleration of 0.5m/s2, and B makes a uniform motion in the same direction at a speed of 5m / s 200m in front of A. Q: (1) When will a catch up with B? How fast does a catch up with B? How far is a from the starting point? (2) When is the maximum distance between a and B in the process of catching up? What is the distance?

(1) When a catches up with B, their displacement difference is x0 = 200m,   X a = x0 + X B. suppose a catches up with B after time t, then x a = 12a a T2, x B = v b T. according to the catch-up conditions, 12a T2 = v b t + 200, the solution is t = 40   S or T = - 20   S (rounded). At this time, the speed of a V A = a a

There are two cars of Party A and Party B on the straight highway. Party A starts driving at a standstill with an acceleration of 0.5m / S ^ 2, B moves at a uniform speed of 5m / s at 200m in front of A. Q: (1) when does a catch up with B? How fast does a catch up with B? How far is a from the starting point at this time? (2) when is the maximum distance between a and B in the process of catching up? What is the distance? Another problem is that the skydiver performs low altitude skydiving. He is still at 224m above the ground and begins to do free fall in the vertical direction. After a period of time, immediately open the parachute and descend at an average acceleration of 12.5m / S ^ 2. For the safety of the athlete, it is required that the maximum landing speed of the athletes shall not exceed 5m / S (G is 10m / S ^ 2) (1) What is the minimum height from the ground when the athlete opens the umbrella? How high is it equivalent to falling freely when landing on the ground? (2) What is the shortest time for an athlete to be in the air? Go back and add points

Question 1: v b * t + 200 = a * T ^ 2 / 2; 5*t+200=0.5*t^2/2； T1 = 40, T2 = - 20 (rounded off); After 40s, the final speed VT = a * t = 0.5 * 40 = 20m / S; Displacement of a = a * T ^ 2 / 2 = 0.5 * 40 * 40 / 2 = 400m; When the relative velocity is equal to 0, their distance is the largest: v b = a a * t; 5=0.5*t...