The car starts to do uniform acceleration motion from static, and then does uniform motion after reaching the speed of V = 10m / s. After the total motion time is 12s, the distance the car passes is 80m The car starts to do uniform acceleration motion from static state, and then does uniform acceleration motion after reaching the speed of V = 10m / s. After the total motion time is 12s, the distance that the car passes is 80m. How long is the acceleration motion? What is the acceleration at starting?

The car starts to do uniform acceleration motion from static, and then does uniform motion after reaching the speed of V = 10m / s. After the total motion time is 12s, the distance the car passes is 80m The car starts to do uniform acceleration motion from static state, and then does uniform acceleration motion after reaching the speed of V = 10m / s. After the total motion time is 12s, the distance that the car passes is 80m. How long is the acceleration motion? What is the acceleration at starting?

If the acceleration is a, the distance is s, and the acceleration time is t, then the uniform driving time is (12-t)
From v = AT1 to 10 = at
And S = 1 / 2at2 + V (12-t), that is, 80 = 1 / 2 * 10t + 10 * (12-t)
The solution is: T = 8s, a = 5 / 4 m / S2

The car is moving forward on the straight road at the speed of 10m / s. there is a bicycle moving in the same direction at the speed of 4m / s right in front of it. The car immediately turns off the accelerator to do the uniform speed change movement of a = - 6m / S2. If the car just can't touch the bicycle, the size of X is ()
A. 9.67mB. 3.33mC. 3mD. 7m

The time required for the car to decelerate to 4m / s t = V2 − V1a = 1s. At this time, the displacement of the car X1 = v1t + 12at2 = 10 × 1-12 × 6 × 1 = 7m. The displacement of the bicycle x2 = v2t = 4 × 1 = 4m. If the car just can't touch the bicycle, there are: x2 + x = x1, so x = 3M. So C is correct, a, B, D are wrong