Why is the displacement formula s = v0t + 1 / 2at ^ 2
It's interpreted according to the V-T image
The displacement is expressed by area
S = rectangle + triangle (triangle is the displacement of the image when the initial velocity is 0)
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- 1. S = v0t + 1 / 2at square The physical meaning of the expression and the meaning of each item
- 2. How does the physical formula s = 1 / 2at come from?
- 3. What is the difference or internal relationship between acceleration and velocity? What is the direction of velocity in S-T or V-T image?
- 4. Why is acceleration in meters per second Why does the formula for calculating acceleration use half of the difference between the final velocity and the initial velocity
- 5. What does the second power of s represent?
- 6. The relationship between the displacement of a particle and time is: S = 4T + 2 times the square of T, the units of S and T are meter and second respectively, The relationship between displacement and time is: S = 4 * t + 2 * t * t Therefore, the relationship between velocity and time is v = DS / dt = 4 + 4 * t So the initial velocity of the particle is: V0 = 4 + 4 * 0 = 4m / s The relationship between acceleration and time is a = DV / dt = 4 So the acceleration of particle is a = 4m / S2 The relationship between the velocity and time is v = DS / dt = 4 + 4 * t Can you tell me how to push it out? We haven't learned derivative yet Please tell me v=ds/dt=4+4*t My feeling is that v = DS / dt = 4 + 2 * t is right
- 7. The relationship between the displacement of a particle and time is s = 4T + 2T square. If the units of S and T are m and s respectively, what are the initial velocity and acceleration of the particle
- 8. The relationship between the displacement of a particle and time is: S = 4T + 2t2, the units of S and T are m and s respectively,
- 9. The relationship between the displacement of a particle and time x = 4T + 2T ^ 2 x T unit is meter and second, then the initial velocity and acceleration of the particle are x=4t+2t^2
- 10. The relationship between the displacement of a particle and time is x = 4T + 2t2 + 3. If the units of X and T are meter and second respectively, the initial velocity and acceleration of the particle are () A. 4m / s and 4m / S2 b.0 and 4m / s2 C. 4m / s and 2m / S2 d4m / s and 0
- 11. S = v0t + 1 / 2at square s = v0t-1 / 2at square what's the physical formula? Why didn't my teacher teach it,
- 12. S = VT + 1 / 2at ^ 2 VT ^ 2-v0 ^ 2 = 2As how can these two formulas be derived Ask for guidance
- 13. The difference between the physical displacement formula s = v0t + 1 / 2at ^ 2 and S = vt
- 14. S = VT + 1 / 2at & sup2; if the time unit is h and the displacement unit is km, how will the formula change?
- 15. In the high school physics formula, there is a formula for acceleration: the square of S = at. How can this formula be deduced?
- 16. In the formula s = VT + 1 / 2at & # 178;, when t = 1, s = 13; when t = 5, Ms = 225, find the value of s when t = 2
- 17. The displacement of an object from the top of the inclined plane to the bottom of the inclined plane is S1 in the first three seconds and S2 in the last three seconds It is known that S1 + S2 = 1.2m, Si: S2 = 3:7 In one step, I can't figure out why v = V 0 + at = 0.28 + 0.08 * 1.5, that 1.5 is. Isn't V 0 the initial velocity? Why can we use the average velocity instead Calculate S1, S2, then a = 0.08, then calculate the average speed in the next 3 seconds = 0.28, then calculate the final speed, calculate t according to v = at, and then substitute t into S = 1 / 2at ˇ 2 to calculate s = 1m, but when calculating the final speed, it is how to calculate v = V0 + at = 0.28 + 0.08 * 1.5, and 0.28 is how to substitute the average speed into the initial speed
- 18. The object from the top of the inclined plane from static to the bottom of the inclined plane, the first 3 seconds after the displacement S1 S2 S1 + S2 = 1.2m S1: S2 = 3:7, calculate the length of the inclined plane
- 19. It is known that S2 + S1 = 1.2m, S1: S2 = 3:7, and the total length of the inclined plane is 1.2m?
- 20. The displacement of an object is S1 in the first 3 seconds and S2 in the last 3 seconds The displacement of an object in the first 3 seconds is S1, and that in the last 3 seconds is S2. Given s2-s1 = 6m, S1: S2 = 3:7, the total length of the inclined plane can be calculated The slide is accelerating. Let the length of slope be l and the total time be t L=a*t^2 / 2 S1=a *3^2 / 2 S2=L-[ a*( t -3)^2 / 2 ]=3at-4.5a 3/7=(a *3^2 / 2)/(3at-4.5a) T = 5 seconds 6=(3at-4.5a)-(a *3^2 / 2) a=1m/s^2 L = 12.5A = 12.5m