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A vehicle traveling along a flat and straight road has a speed of 36km / h. The acceleration obtained after braking is 4m / S2. Find: (1) Speed at the end of 3S after braking; (2) The speed at which the vehicle coasts half the distance from the start of braking to the stop
(1) 36km / h = 10m / s according to the speed time formula, the time required for the vehicle to reduce the braking speed to zero t = 0 − v0a = − 10 − 4S = 2.5s ＜ 3S. Then the speed at the end of 3S after braking is zero. (2) if the speed at half the sliding distance is V, then V2 − V02 = 2A • X20 − V2 = 2A • X2, the simultaneous solution of two formulas is v = V022
If the vehicle runs at a constant speed on a straight highway at the speed of 36km / h and the acceleration during braking is 4m / S2, the displacement of the vehicle within 5S after braking is______ m．
According to v = V0 + at, the time required when the vehicle speed decreases to 0 is t = 0 − 36000
three thousand and six hundred
−4s＝2.5s
That is, the car stops moving after 2.5s, so the displacement of the car within 5S after braking is x = v0t + 1
2at2＝10 × 2.5−1
two × four × 2.52m=12.5m
So the answer is: 12.5
The speed of a vehicle traveling along a flat and straight road is 36km / h, and the acceleration obtained after braking is 4m / S2. From the beginning of braking to the stop, Taxi distance
36km/h=10m/s
Braking time: T = (VT VO) / a = (0-10) / (- 4) = 2.5m/s ^ 2
Sliding distance: S = VO * t / 2 = 10 * 2.5 / 2 = 12.5m
For a vehicle traveling eastward at the speed of 18m / s on a straight highway, make a uniform deceleration movement after braking, and advance 36m within 3S. Calculate the acceleration and square of the vehicle The speed at the end of the 6th s after the car starts braking, and the distance the car passes from the beginning of braking to the complete stop
According to the uniform deceleration formula: S = VT - 1 / 2 at ² Yes: 36 = 18 × 3-1/2 a × three ² Solution: a = 4 m / S ² (the acceleration direction is the same as the braking force direction, westward) according to vt VO = - at, when the final speed is zero, the time from braking to stopping can be calculated: 18 = 4tt = 4.5 seconds, visible for 6 seconds
An object moves in a straight line with uniform acceleration. The initial velocity is 0.5m/s. The displacement in the 7th s is 4m more than that in the 5th S. calculate the displacement of the object in the 5S The answer is 27.5m 9m
Displacement in the 7th second: S7 = 7v0 + 1 / 2A * 7 ^ 2 (1) displacement in the 6th second: S6 = 6v0 + 1 / 2A * 6 ^ 2 (2) displacement in the 5th second: S5 = 5v0 + 1 / 2A * 5 ^ 2 (3) displacement in the 4th second: S4 = 4v0 + 1 / 2A * 4 ^ 2 (4) (1) - (2): displacement in the 7th second #7 = V0 + 1 / 2A (7 ^ 2-6 ^ 2) (5) (3) - (4): displacement in the 5th second #5
An object moves in a straight line with uniform acceleration, and the initial velocity is 0.5m/s. The displacement in the 7th s is 4m more than that in the 5th S. find: (1) the acceleration of the object; Xm-Xn=(m-n)aT ² How does this formula determine t = 1
m=7
n=5
T = 1 (because the 7S and 5S of course refer to one second)
So: 4 = (7-5) a * 1 ^ 2
Solution: a = 2m / S ^ 2
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