On the problem of uniform speed change linear motion in senior one physics If the stop time is 1 min, the acceleration caused by braking is 30 cm / (S2), and the acceleration caused by starting is 50 cm / (S2)?

On the problem of uniform speed change linear motion in senior one physics If the stop time is 1 min, the acceleration caused by braking is 30 cm / (S2), and the acceleration caused by starting is 50 cm / (S2)?

The normal running speed of the train: v = 54km / h = 54 × 10 & sup3; m / 3600s = 15m / s. from the formula 2As = V2 & sup2; - V1 & sup2;, the braking distance can be obtained: S1 = V & sup2; / 2A1 = 15 & sup2; / (2 × 0.3) = 375 (m); the starting acceleration distance can be obtained: S2 = V & sup2; / 2A2 = 15 & sup2; / (2 × 0.5) = 225 (m)

There is an ejection system on the US "Kennedy" aircraft carrier to help the aircraft take off. It is known that the acceleration produced by the "f-a15" fighter when accelerating on the runway is 4.5m/s2, and the take-off speed is 50M / s. if the aircraft takes off when taxiing 100m, the initial speed of the ejection system must be ()
A. 30m/sB. 40m/sC. 20m/sD. 10m/s

Let the initial velocity of the aircraft be V0, the acceleration a = 4.5m/s2, the displacement x = 100m, and the final velocity v = 50M / s of the aircraft are known. This problem does not involve the time of the motion of the object, then the displacement time formula of the uniform variable speed linear motion: v2-v02 = 2aX is used to solve: V0 = 40m / s, so select: B

After turning off the engine, the car taxis into the station in a straight line with uniform deceleration. It is known that when taxiing 120 m, the speed is reduced to half of the original speed, and then taxis for 8 s at rest. The speed and taxiing distance of the car when turning off the engine are calculated

If the initial velocity is V0 and the acceleration is a, then the velocity is exactly half of the initial velocity: v0-v02 = at12ax1 = (V02) 2-v02 in the process; 0-v02 = at2 in the second half; the data solution: V0 = 20m / s, a = - 1.25m/s2, the second displacement: x2 = (V02) 22 & nbsp; a = 40m, so the total displacement