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On a straight highway, the speed of a car is 15m / s. It starts braking at a certain time. Under the action of resistance, the car moves at an acceleration of 2m / S2. How far does the last car leave the starting braking point 10 s after braking?
If the time from braking to stopping is t0, then v = V0 + at0 = 0:
t0=−v0
a＝15
2＝7.5s
If the vehicle stops moving 7.5S after braking, the displacement of the vehicle within 10s after braking is equal to the displacement within 7.5S after braking, and the size is:
x=v0t0
2＝15 × seven point five
2m=56.25m
Answer: 10s after braking, the distance between the last car and the starting braking point is 56.25m
The speed of a car is 10m / s. It starts braking at a certain time. Under the action of resistance, the car decelerates at a uniform speed with an acceleration of 2m / S2 Ask: how far is the car from the braking point after 10s?
The speed of a car is 10m / s. It starts braking at a certain time. Under the action of resistance, the car decelerates at a uniform speed with an acceleration of 2m / S2. It stops in 5 seconds! So the distance S = (10 + 0) / 2 × 5 = 25m! (based on average speed multiplied by time)
On a straight highway, when the vehicle starts braking at a certain time at a speed of 15m / s, under the action of resistance, the vehicle decelerates at an acceleration of 2m / s, what is the displacement within 10s after braking?
Because of deceleration, the acceleration is negative, that is, a = - 2m / s. method 1: substitute the formula v = V0 + at into the data; Get the time t = 7.5S. From the formula s = v0t + 1 / 2at2, (initial velocity multiplied by time plus half of the acceleration multiplied by time square), get s = 56.25m. Method 2: deduce the formula: v2-v02 = 2As, (velocity square
When the car moves in a uniform straight line at the initial speed of 10m / s on a straight highway, it finds something in front and brakes, and the acceleration is 2m / s ² Then: {1} what is the speed of the car after 3 seconds? What is the speed of {2} after 5 seconds? What is the speed of {3} after 10 seconds?
（1）10-2*3=4
（2）10-2*5=0
(3) 10-2 * 5 = 0, because the car has stopped after 5 seconds, so after 10 seconds, the car certainly stops, and the speed is 0
The speed before braking is 20m / s, and the acceleration obtained by braking is 2.5m/s2 (for analysis and answer) (1) Displacement in the second second second (2) The distance of the car sliding in the last 2S before standstill (3) The sliding distance within 10s after the start of vehicle braking
(1) Speed in the first second V1 = 20-2.5 * 1 = 17.5m/s
s=v1t-at^2/2=17.5-1.25=16.25m
(2）s=2.5*2^2/2=5m
(3) Braking time t = 20 / 2.5 = 8s
Sliding distance S = 2.5 * 8 ^ 2 / 2 = 80m
The car moves in a uniform straight line at a speed of 20m / s, and the acceleration after braking is 5m / S2, then the ratio of the displacement passed by the car within 2S after braking to that within 6S after braking is () A. 1：1 B. 3：1 C. 4：3 D. 3：4
The time t0 = 0-v0a = 0-20-5s = 4S required for the vehicle to stop after braking. Then the displacement within 2S after braking x = v0t + 12at2 = 20 × 2-12 × five × 4m = 30m. The displacement within 6S after braking is equal to the displacement within 4S after braking, then x '= 0-v022a = - 400-10m = 40m. Then the displacement of the vehicle passing within 2S and 6S after braking
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