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The speed before braking is 5m / s, and the acceleration of 0.4m/s is obtained when braking. Find: (1) (2) (3) write the specific steps The speed before braking is 5m / s, and the acceleration of 0.4m/s is obtained when braking Find: (1) the sliding distance within 20s after braking; (2) The time from opening the brake to when the displacement of the vehicle is 30m; (3) Sliding distance of vehicle within 2.5s of standstill ***Be sure to write down the specific steps
1、 If the acceleration is 0.4m/s, the vehicle has stopped moving at 12.5s, s = vt-1 / 2at square, s = 62.5-31.25 = 31.25m
2、 2As = V0 square - V square, then v = 1m / s, v = VO at, t = 10s
3、 The problem is not clear, do not understand, is to stop moving? Or the first 2.5s?
If it stops, it must be 0. If it starts, it is s = vt-1 / 2at square, s = 11.25m
A car runs along a straight road at a speed of 20m / S; When the vehicle brakes at an acceleration of 5m / S2, the braking distance is () A. 40m B. 20m C. 100m D. 4m
The final speed of braking is zero, according to 2aX = V2 − V02:
x=0−400
−10＝40m
Therefore: a
The car runs at a constant speed on the road at a speed of 10 meters per second. The speed changes from 2 seconds after braking to 6 meters per second. Calculate the distance 2 seconds after braking and the acceleration when braking Have the ability to calculate the time taken to advance 9 meters; Forward distance after braking
Known: T = 2S VO = 6m / s VT = 10m / s VT = VO + AT1. A = (6-10) / 2 = - 2m / S ² 2. Distance at this time s = VOT + at ²/ 2 = 20-4 = 16m -------------------------------- advance 9m: the formula is the same as above (2.) 9 = 10t-t ² T ²- 10T+9=0 T ²- 10T+25-1...
The speed of the car before braking is 5 meters per second, and the braking obtains an acceleration of 0.4 meters per second. Calculate: (1) the sliding distance of the car 20 seconds after braking. (2) the time of sliding 30 meters after braking
(1) Calculate stop time t
0=v0+aT
0=5-0.4T
Because t = 12.5s ＜ 20s
So 0-v0 ^ 2 = 2aX
-25=-0.8x
x=31.25m
(2)s=v0t+1/2at^2
30=5t+0.2t^2
t=5s
The speed before braking is 20 meters per second, and the acceleration obtained by braking is 2 meters per square second Find (1) the sliding distance within 20 seconds after the start of vehicle braking (2) The time from the start of braking to the car displacement of 30 meters (3) The sliding distance of the vehicle within 2.5 seconds before standstill
1）v^2=2as,s=v^2/(2a)=20^2/(2*2)=100m
2) Calculate the area enclosed by V-T diagram line, and set t:
(v0+v0-at)*t/2=s1,(20+20-2t)*t/2=30,
Because t (max) = V / a = 20 / 2 = 10s,
‡ t = 10 root sign 70
3) V = at = 2 * 2.5 = 5m / s can also be calculated inversely by using V-T diagram line,
s=(5+0)*2.5/2=6.25m
The car moves in a straight line at a constant speed of 20 meters per second. After braking, the acceleration obtained is 5 every quadratic second, then the car slides within 2 seconds after braking The displacement of is
S=Vot-1/2at^2=20x2-1/2x5x2^2=30m
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