Question

请教高中物理问题，求教高手，期望有分析过程

Answer 1

因为打点记录器的周期是 0.02S，图上两点之间是 5个打点周期，即 T= 0.1S。

1. Sab = V0*T+1/2*a*T^2 = 0.0210 m

2. Sac = V0*(2T)+1/2*a*(2T)^2 = 2V0*T + 2a*T^2 = 0.0620 m

3. Sad = V0*(3T)+1/2*a*(3T)^2 = 3V0*T + 4.5 *a*T^2 = 0.1230 m

4. Sae = V0*(4T)+1/2*a*(4T)^2 = 4V0*T + 8a*T^2 = 0.2040 m

5. Saf = V0*(5T)+1/2*a*(5T)^2 = 5V0*T + 12.5a*T^2 = 0.3050 m

    

   Sbc = Sac - Sab = V0*T + 1.5a*T^2 = 0.0410 m

   Scd = Sad - Sac = V0*T + 2.5a*T^2 = 0.0610 m

   Sde = Sae - Sad = V0*T + 3.5a*T^2 = 0.0810 m

   Sef  = Saf - Sae  = V0*T + 4.5a*T^2 = 0.1010 m

6. Scd - Sbc = a*T^2 = a*(0.1)^2 = 0.01*a = 0.0200 m，a = 2.00 m/s

   Sde - Scd = a*T^2 = 0.01*a = 0.0200 m，a = 2.00 m/s

   Sef - Sde = a*T^2 = 0.01 *a = 0.0200 m，a = 2.00 m/s

   所以，根据统计结果，取平均值，可以得到加速度 a = 2.00 m/s

Answer 2

每二个计数点之间的时间间隔为T=0.1s
由逐差法公式
a1=(s2-s1)/T^2
a2=(s3-s2)/T^2
a3=(s4-s3)/T^2
a4=(s5-s4)/T^2
a=(a1+a2+a3+a4)/4=(S4-S1)/4T^2=(10.10-2.10)*10^-2/4*0.01=8.0/4=2.0m/s^2