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On the straight road, there are two cars moving in the same direction at the same place. Car a moves in a straight line at a speed of 10m / s, and car B moves in a straight line at a speed of 1m / S2 from a standstill. Question: (1) when will car a and car B meet again after departure? (2) When is the farthest distance between the two cars before they meet again? What's the longest distance?
(1) Let a meet again after time T. at this time, the displacement of car a is X1 = v1t = 10t, and the displacement of car B is x2 = 12at2. When meeting again, the displacement is equal, 12at2 = vt. substituting the data, t = 20s. So two cars meet again after 20s. (2) when the speed of two cars is equal, the distance is the farthest
Car a runs at a constant speed on a straight road at a speed of 10m / s. car B moves in a straight line at a speed of 4m / s in parallel with car a in the same direction. Car a starts to brake at an acceleration of 0.5m/s2 after passing by car B. the timing starts from car a's braking
(1) The time for car B to catch up with car A. my algorithm is to first find out the time for car a to stop moving = 20s, calculate the distance for car a to move within 20s = 100m, and then 100 / 4 = 25s, that is, the time for car B to catch up with car a is 25s, but the answer is 24s. Who can tell me what's wrong with my idea?
Wrong, because it's not when car a is still
When car B and car a have the same speed, they can catch up
Or they catch up when they move the same distance
You set time t to catch up
Then 4T = 10t-0.5 * 0.5T ^ 2
The solution is t = 24s
Car a runs on a straight road at a speed of 10m / s, and car B moves in a straight line with car a at a speed of 4m / s. car a starts to brake at an acceleration of 0.5m/s2 when passing by car B. the timing starts from car a's braking. The following are calculated: (1) the maximum distance between two cars before car B overtakes car a; (2) the time it takes car B to overtake car a
(1) Suppose that the starting speed of car a is V1 and that of car B is v2. Before car B overtakes car a, the distance between the two workshops is the largest when the speed of the two cars is equal. Suppose that the time experienced at this time is T1, then from V1 = V2 + AT1, T1 = V1 − V2a = 10m / s − 4m / s0.5m/s2 = 12s, then the displacement of car a is v1t1 + 12at12 = 10 × 12 − 12 × 0.5 × 122m = 84m. X1 = v1t1 + 12at12 = 10 × 12-12 × 0.5 × 122 = 84m The displacement of vehicle B is x2 = v2t1 = 4 × 12m = 48m, so the maximum distance between the two workshops is △ x = x2-x1 = 84-48m = 36m. (2) if the passing time t is set for vehicle B to catch up with vehicle a, and the displacement of the two vehicles is the same, then t = 24s is obtained by substituting v2t = v1t + 12at2 into the data solution. The time t ′ = V1a = 20s for vehicle a to stop braking According to the data, 1001 = 4T ″, t ″ = 25s. A: (1) before car B overtakes car a, the maximum distance between the two cars is 36m. (2) it takes 25s for car B to overtake car a
Car a runs on a straight road at a speed of 10m / s, and car B moves in a straight line with car a at a speed of 4m / s. car a starts to brake at an acceleration of 0.5m/s2 when passing by car B. the timing starts from car a's braking. The following are calculated: (1) the maximum distance between two cars before car B overtakes car a; (2) the time it takes car B to overtake car a
(1) Suppose that the starting speed of car a is V1, and that of car B is v2. Before car B overtakes car a, the distance between the two workshops is the largest when the speed of the two cars is equal. Suppose that the time experienced at this time is T1, then from V1 = V2 + AT1, T1 = V1 − V2a = 10m / s − 4m / s0.5m/s2 = 12s, then the displacement of car a is v1t1 + 12At
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