When an object moves in a straight line with uniform deceleration, the initial speed is 10 meters per second and the acceleration is 1 meter per second square, what is the average speed of the object 2 seconds before it stops?

Using the inverse process, the average speed of the object 2 seconds before the stop, the initial speed is 0, and the acceleration is 1 meter per second. The average speed of the object 2 seconds before the uniformly accelerated linear motion, that is, the speed at the end of the first second (intermediate time) = 1m / s

An object moves in a straight line with uniform deceleration at an initial speed of 20 meters per second. The acceleration is 5 seconds per meter square. 1 find the object to stop after a few seconds 2 find the displacement of the object at the end of 6 seconds (specific process)

0=20m/s+（-5m/s2）*t,t=4s； The displacement at the end of 6S is the same as that at the end of 4S: x = 0.5 * 5 * 4 ^ 2 = 40m

An object moves in a straight line with uniform deceleration at the initial speed V0 = 12m / s until it stops, and the acceleration is 2m / quadratic second Find: 1) the velocity of the object at the end of 3S; 2) the displacement and average velocity of the object within 4S 3) Average velocity in the 4th s 4) average velocity within the first 20m displacement of the object

Think of it as a vertical throw up! Before you can solve the answer! I only provide ideas! (the premise is that you have to recite the relevant formula) 1. First calculate how long it takes when the speed is equal to 0 (if it is greater than 3 seconds, use VT = VO + A * t: A is a vector with size and direction. Pay attention to the direction here. I guess it should not be less than 3 seconds) 2. The same

A train moves in a straight line with uniform acceleration at an initial speed of 2 meters per second and an acceleration of 0.5 meters per square second. Find: (1) what is the initial speed of the train at the end of the third second? (2) What is the average speed in the first four seconds? (3) What is the displacement in the fifth second? (4) What is the displacement in the second 4 seconds?

1、V=V0+at1=2m/s+0.5 × 3m/s=3.5m/s
2. Displacement in the first four seconds S2 = v0t2 + 1 / 2at2 ^ 2 = 2 × 4m+1/2 × zero point five × 4^2m=9m
Average speed v = S2 / T2 = 9 / 4m / S = 2.25m/s
3. Displacement in the first five seconds S3 = v0t3 + 1 / 2at3 ^ 2 = 2 × 5m+1/2 × zero point five × 5^2m=16.25m
Then the fifth second displacement Δ S=S3-S2=7.25m
4. Displacement in the first eight seconds S4 = v0t4 + 1 / 2at4 ^ 2 = 2 × 8m+1/2 × zero point five × 8^2m=32m
So the displacement in the second 4 seconds Δ S' = S4-S2=23m

A train moves in a straight line with an initial speed of 2 meters per second and an acceleration of 0.5 meters per square meter. What is the displacement in the fifth second

The problem is the fifth second
Then subtract the displacement of the first five seconds from the displacement of the first four seconds
2 * 5 + 1 / 2 * 0.5 * 5 ^ 2 = 10 + 25 / 4 in the first 5 seconds
First four seconds 2 * 4 + 1 / 2 * 0.5x4 ^ 2 = 8 + 16 / 4
4.25

A train moves in a straight line with an initial speed of 2 meters per second and an acceleration of 0.5 meters per square second. Find out the speed of the train at the end of the third second

Speed v = V0 + at = 2 + 0.5 * 3 = 3.5 meters per second

When an object moves in a straight line with uniform deceleration, the initial speed is 10 meters per second and the acceleration is 1 meter per second square, what is the average speed of the object 2 seconds before it stops?