Two objects a and B move in a straight line in the same direction when they pass a certain intersection at the same time. A moves at a constant speed of 4 meters per second. B moves in a straight line with a constant acceleration of 2 m / S ^ 2 from the intersection. Q: how long does it take for them to start timing when they pass the intersection, The first time they met? How far away from the intersection when they met? When was the maximum distance between the two objects before they met? What is the maximum distance?

Two objects a and B move in a straight line in the same direction when they pass a certain intersection at the same time. A moves at a constant speed of 4 meters per second. B moves in a straight line with a constant acceleration of 2 m / S ^ 2 from the intersection. Q: how long does it take for them to start timing when they pass the intersection, The first time they met? How far away from the intersection when they met? When was the maximum distance between the two objects before they met? What is the maximum distance?

(1) For the first time, the displacement is the same. A moves in a straight line at a uniform speed of 4 m / s, so the displacement is 4 T
Displacement of B: according to the formula x = v0t + 1 / 2A (T) ^ 2, the displacement is T ^ 2, and a formula: 4T = T ^ 2 is obtained
The solution is t = 4
T = 4S v = 4m / s for X
x=vt=16m.
(3) Before the speed of B catches up with a, the distance becomes larger and larger. So when the speed of B is 4m / s, the distance is the largest
According to v = V0 + at, we get 4 = 0 + at, at = 4, and because a = 2m / S ^ 2, t = 2S
(4) T = 2S, v = 4m / s, so x = 8m
Is my answer good?

A and B move in a straight line at a constant speed. The speed of a is twice that of B. if the ratio of the distance between a and B is 2:1, the time ratio of a and B is ()
A. 1：1B. 2：1C. 4：1D. 1：2

The ratio of t a t b = s a V a s b v b = s a v b s b v a = 21 × V B 2 & nbsp; v b = 11; so a

A horizontal conveyor belt is 20 m long and moves at a constant speed of 2 m / s. now, a small object is gently placed on one end of the conveyor belt to make it move from a static state. If the dynamic friction coefficient between the object and the conveyor belt is 0.1, the following results are obtained: (1) the time required for the small object to reach the other end of the conveyor belt; (2) the trace left by the relative movement of the object on the conveyor belt

F = μ n = ma n = mg a = μ G
a=（△V/△t)
Acceleration a = 1m / S ^ 2 T1 = (△ V / △ a) = V / a = 2S s = 1 / 2 × 1 × 4 = 2m. At this time, the conveyor belt moves 2 × 2 = 4m
t2=(20-2)/2=9s
The time required is T1 + T2 = 11S, and the relative trace left is 2m

A and B are moving in a straight line at a constant speed. A is 20 meters in front of B and two seconds earlier than B. A's speed is 2 m / s. B's speed is 10.8 km / h. B can catch up with B in many times

Let B overtake a in T seconds
First, unify the speed of unit B: 10.8km/h = 10.8 / 3.6m/s = 3m / s
A is 20 meters ahead of B and two seconds ahead of B = > 20 + 2 * 2 + 2T = 3T
The solution is t = 24s
A: after 24 seconds, B overtakes a